Volume 9, No. 3, May-June 2018
International Journal of Advanced Research in Computer Science
RESEARCH PAPER
Available Online at www.ijarcs.info
ISSN No. 0976-5697
SOME FIXED POINT THEOREMS IN CONE RANDOM METRIC SPACE USING
RANDOM OPERATORS
Gourish Parashar
Department of Mathematics TIT &
Science Bhopal and Research Scholar
M.G.C.G.V. Chitrakoot (M.P.), India
Anil Agrawal
Department of Mathematics
M.G.C.G.V. Chitrakoot (M.P.), India
Manoj Kumar Shukla
Department of
Mathematics,
Institute
for
Excellence
in
Higher
Education
,
Bhopal
,
(M.P.), India
Abstract: Some fixed point theorems in cone random metric space by use of random operators with different contractions are proved which are generalizations of contraction mappings considered by various researchers.
Keyword: Cone Random Metric Space, Random operator,Random fixed point, Implicit relation.
AMS CLASSIFICATION: 47H10, 54H25.
1.INTRODUCTION
Random fixed point theorems for contraction mappings in polish spaces and random fixed point theorems are of fundamental importance in probabilistic functional analysis. The Random fixed point theorems were initially studied by the Prague school of Probabilistic. In 1955 it was Spacek [28] who introduced this topic further Hans[11], [12], extended the work of Spacek. Subsequently Kannan, R. [16], Bharucha-Ried [8], Itoh [15] proved several random fixed point theorems and gave their applications to random differential equations in Banach spaces. Random coincidence point theorems and random fixed point theorem are stochastic generalization of classical coincidence point theorems and classical fixed point theorems. Sehgal and Singh [26], Papageorgiou [22], Rhoades Sessa Khan [25] and Lin [19] have proved differential stochastic version of well known Schauderβs fixed point theorem. Then Beg, I. and Azam [2], Beg and Shahzad [3], [4], [5], studied the structure of common random fixed points and random coincidence points of a pair of compatible random operators. Further the cone metric space was defined by Huang and Zhang [14], they generalized the concept of metric spaces, replacing the set of real numbers by an ordered Banach space; Huang and Zhang also described the convergence of sequences and introduced the notion of completeness in cone metric spaces. They have proved some fixed point theorems of contractive mappings on complete cone metric space with the assumption of normality of a cone. There exist a lot of work involving fixed point used the Banach contraction principle. This principle has been extended kind of contraction mappings considered by various authors[1],[6],[7],[9], [13] and [27]. Recently, in 2008 various authors such as Dhagat, V. B, Sharma, A. K, Bhardwaj, R. K. [10], have studied Fixed point theorem for
Random operators in Hilbert Spaces, this result was a generalization of the results of Huang and Zhang also they studied fixed point theorems for normal and non-normal cones. Recently Rezapour et.al.[23] ,[24] Pathak, H. K. and Shahzad, N.[17] did some notable work in quashi contraction maps and cone metric spaces with contractive mapping. Then in 2010 Sumitra et. al. [29] and Mehta Smrati [20] in 2011 extended the previous results in different spaces. Here, we have prove some fixed point theorems in cone random metric space by using random operators with different contractions.
2. PRELIMINARIES
Definition 2.1: Let (πΈ, π) be a topological vector space and
π a subset of πΈ, π is called a cone if
1. π is non-empty and closed, π β {0},
2. For π₯, π¦ β π and π, π π π βΉ ππ₯ + ππ¦ β π
where π, π β₯ 0
3. If π₯ β π and β π₯ β π βΉ π₯ = 0
For a given cone π β πΈ, a partial ordering β€ with respect to
π is defined by π₯ β€ π¦ if and only if π¦ β π₯ β π,π₯ < π¦ if π₯ β€ π¦ and π₯ β π¦,while π₯ βͺ π¦ will stand for π¦ β π₯ β πππ‘π
denotes the interior of π.
Definition 2.2: Measurable function: Let (β¦, β) be a measurable space with β a sigma algebra of subsets of β¦ and π a non-empty subset of a metric space π = (π, π). Let
2π be the family of all non-empty subsets of π where
πΆ(π) the family of all nonempty closed subsets of π. A mapping πΊ βΆ β¦ β 2π is called measurable if, for each open
πΊβ1(π) β β, where πΊβ1(π) =
{π β β¦ βΆ πΊ(π) β© π β β }.
Definition 2.3: Measurable selector [18]: A mapping π βΆ β¦ β π is called a measurable selector of a measurable mapping πΊ βΆ β¦ β 2π if π is measurable and π(π) β
πΊ(π) for each π β β¦.
Definition 2.4: Random operator [21]: Mapping π βΆ β¦ Γ π β π is said to be a random operator if, for each fixed π₯ β π, π(. , π₯) βΆ β¦ β π is measurable.
Definition 2.5: Continuous Random operator [21]: A random operator π βΆ β¦ Γ π β π is said to be continuous random operator if, for each fixed π₯ β π, π(. , π₯) βΆ β¦ β π is continuous.
Definition 2.6: Random fixed point: A measurable mapping π βΆ β¦ β π is a random fixed point of a random operator π βΆ β¦ Γ π β π if π(π) = π(π, π(π)) for each
π β β¦.
Definition 2.7: Let π be a nonempty set and the mapping
π βΆ β¦ Γ π β π and Pβ π be a cone, π β β¦ be a selector, satisfies the following conditions:
2.7(π) π(π₯(π), π¦(π)) > 0β π₯(π), π¦(π) β β¦ Γ π βΊ π₯(π) = π¦(π)
2.7(π) π(π₯(π), π¦(π)) = π(π¦(π), π₯(π)) β π₯, π¦ β π, π β β¦ and π₯(π), π¦(π) β β¦ Γ π
2.7(π) π(π₯(π), π¦(π)) = π(π₯(π), π§(π)) + π(π§(π), π¦(π))β π₯, π¦ β π and π β β¦ be a
selector.
2.7(π) for any π₯, π¦ β π, π β β¦,
π(π₯(π), π¦(π)) is non-increasing and left continuous in πΌ. Then π is called cone random metric in π and (π, π) is a cone random metric space.
Definition 2.8: Implicit Relation
Let Ξ¦ be the class of all real-valued continuous functions βΆ (π +)3β π + non-decreasing in the first argument and
satisfying the following conditions:
π β€ π(π¦, π₯ + π¦, π¦) or π₯ β€ π(π¦, π₯, π₯) or π₯ β€ π(π¦, π₯ + π¦, π₯ + π¦)
there exists a real number 0 < β < 1 such that π₯ β€ βπ¦, for all π₯, π¦ β₯ 0.
Similarly for (π +)5
, Let Ξ¦ be the class of all real-valued continuous functions π βΆ (π +)5β π +
non-decreasing in the first argument and satisfying the following conditions for all
π₯, π¦ β₯ 0.
π β€ π(π¦, π₯ + π¦, π₯ + π¦, π₯ + π¦, π¦) or π β€ π(π¦, 0, π¦, π¦, π₯ + π¦) or π β€ π(π¦, π¦, π₯, π¦, π₯)
there exists a real number 0 < β < 1 such that π₯ β€ βπ¦.
3. MAIN RESULTS
Theorem 3.1: Let (π, π) be a complete cone random metric space and let π be a non-empty separable closed subset of cone metric space π and let π be a continuous random on π
such that for all π β πΊ, π(π, . ): πΊ Γ π β π satisfying contraction
π(π(π₯(π)), π(π¦(π))) β€
π(π(π₯(π), π¦(π)), π(π¦(π), π(π₯(π)) + π(π¦(π), π(π¦(π)),
π(π₯(π), π(π₯(π)) + π(π¦(π), π(π₯(π)),
π(π₯(π), π (π¦(π), π (π¦(π), π(π₯(π)))) β¦ β¦ β¦ (3.1(π))
For all, π¦ β π,π β πΊ. Then π has a fixed point in π.
Proof: For each π₯0(π) β πΊ Γ π and π β₯ 1, let π₯1= ππ₯0
And π₯π+1(π) = π(π₯π(π)) = ππ+1π₯0(π). Then
π(π₯π(π), π₯π+1(π)) = π(π₯πβ1(π)), π(π₯π(π)))
β€ π(π(π₯πβ1(π), π₯π(π)), π(π₯π(π), π(π₯πβ1(π)) +
π(π₯π(π), π(π₯π(π)),
π(π₯πβ1(π), π(π₯πβ1(π)) + π(π₯π(π), π(π₯πβ1(π))
π(π₯πβ1(π), π(π₯π(π)) + π(π₯π(π), π(π₯πβ1(π)))
β€
β (π(π₯πβ1(π), π₯π(π)), π(π₯π(π), π₯π(π)), π(π₯π(π), π₯π+1(π)),
π(π₯π(π), π₯π(π)) + π(π₯πβ1(π), π₯π(π)),
π(π₯πβ1(π), π₯π+1(π)) + π(π₯π(π), π₯π(π)))
β€ π(π(π₯πβ1(π), π₯π(π)), 0 +
π(π₯π(π), π₯π+1(π)), π(π₯πβ1(π), π₯π(π)) + 0,
π(π₯πβ1(π), π₯π(π)) + π(π₯π(π), π₯π+1(π)) + 0)
Therefore by definition (2.8) we have
π(π₯π(π), π₯π+1(π)) β€ β (π(π₯πβ1(π), π₯π(π)))
Similarly
π(π₯πβ1(π), π₯π(π)) β€ β(π(π₯πβ2(π), π₯πβ1(π))
Hence π(π₯π(π), π₯π+1(π)) β€ β (π(π₯πβ1(π), π₯π(π))) β€
β2(π(π₯
πβ2(π), π₯πβ1(π)))
On continuing this process
π(π₯π(π), π₯π+1(π)) β€ βπ(π(π₯0(π), π₯1(π)))
So for π > π
π(π₯π(π), π₯π(π)) β€ (βπ+ βπ+1+ βπ+2+ β― β¦ β¦ +
βπβ1) (π(π₯
0(π), π₯1(π)))
β€ βπ
1ββ(π(π₯0(π), π₯1(π)))
Let 0 βͺ π be given. Choose a natural number π such that
βπ
1ββ(π(π₯0(π), π₯1(π))) βͺ π for every π β₯ π.Thus
π(π₯π(π), π₯π(π)) β€ βπ
1ββ(π(π₯0(π), π₯1(π))) βͺ π for every
π > π β₯ π.
Therefore the sequence {π₯π(π)} is a Cauchy sequence in
πΊ Γ π such that π₯π(π) β π§(π).
Choose a natural number π1 such that
Hence we have
π(π§(π), ππ§(π)) β€ π(π§(π), π₯π+1(π)) +
π(π₯π+1(π), ππ§(π))
= π(π§(π), π₯π+1(π)) + π(ππ₯π(π), ππ§(π))β€
π(π§(π), π₯π+1(π)) +
β (π(π₯π(π), π§(π)), π(π§(π), ππ₯π(π)) + π(π§(π), ππ§(π)),
π(π₯π(π), ππ₯π(π)) + π(π§(π), ππ₯π(π)),
π(π₯π(π), ππ§(π)) + π(π§(π), ππ₯π(π))
β€ π(π§(π), π₯π+1(π)) +
β (π(π₯π(π), π§(π)), π(π§(π), π₯π+1(π)) + π(π§(π), ππ§(π)),
π(π₯π(π), π₯π+1(π)) +
π(π§(π), π₯π+1(π)), π(π₯π(π), ππ§(π)) + π(π§(π), π₯π+1(π))
Taking π ββ we have
π(π§(π), ππ§(π))β€ 0 + β (0, π(π§(π), ππ§(π)) + 0,0 + 0, π(π§(π), ππ§(π)) + 0)
π(π§(π), ππ§(π)) β€ β (0,0 + π(π§(π), ππ§(π)), 0 + 0, π(π§(π), ππ§(π)) + 0))
π(π§(π), ππ§(π)) β€ 0
Thus β (π(π§(π), ππ§(π))) β π. But π(π§(π), ππ§(π)) β π.
Example: Let π = π and π = {π₯ β π: π₯ β₯ 0}, also πΊ = [0,1] and β be the sigma algebra of Lebesgueβs measurable subset of [0,1]. Let π = [0,β) and define mapping as
π: (πΊ Γ π )Γ (πΊ Γ π) β π by π(π₯(π), π¦(π)) = |π₯(π) β π¦(π)|. Then (π, π) is a cone random metric space. Define random operator π from πΊ Γ π to π as π(π₯(π)) = π₯(π)/2. Also sequence of mapping π₯π: πΊ β π is defined
by π₯π(π) = {(1 β (π 2)β 2
)1+1 2β } for every π β πΊ and π β
π. Defined measurable mapping π₯: πΊ β π as π₯(π) = {1 β (π 2)β 2} is fixed point of the space.
Theorem π. π: Let (π, π) be a complete cone random metric space and let π be a nonempty separable closed subset of cone metric space π and let π and π be continuous random operators defined on π such that for π β πΊ, π(π, . ): πΊ Γ π β π satisfying contraction
π(ππ(π₯(π), πππ¦(π))) β€
β (π(π₯(π), π¦(π)), [π (π₯(π), ππ(π₯(π)) +
π(π¦(π), πππ¦(π)))],
π(π₯(π), πππ¦(π)) +
π(πππ₯(π)), π¦(π)) β¦ β¦ β¦ (3.2(π))
For all π₯, π¦ β π,π β πΊ and π > 0. Then π and π has common fixed point in π.
Proof: For each π₯0(π) β πΊ Γ π, let us choose π₯1(π) =
πππ₯
0(π) and π₯2(π) = πππ₯1(π). In general π β₯ 1,
π₯π+1(π) = ππ(π₯π(π)) and π₯π+2(π) = ππ(π₯π+1(π)). Then
π(π₯π+1(π), π₯π+2(π)) = π(ππ(π₯π(π)), ππ(π₯π+1(π)))
β€ β (π((π₯π(π), (π₯π+1(π)), [π((π₯π(π), ππ(π₯π(π)) +
π(π₯π+1(π), ππ(π₯π+1(π))],
[π(π₯π(π), ππ(π₯π+1(π)) + π(ππ(π₯π(π), π₯π+1(π))])
β€ β (π((π₯π(π), (π₯π+1(π)), [π((π₯π(π), (π₯π+1(π)) +
π((π₯π+1(π), (π₯π+2(π))],
[π((π₯π(π), (π₯π+2(π)) + π((π₯π+1(π), (π₯π+1(π))])
β€β (π((π₯π(π), (π₯π+1(π)), [π((π₯π(π), (π₯π+1(π)) +
π((π₯π+1(π), (π₯π+2(π))],
[π((π₯π(π), (π₯π+1(π)))])
Therefore by definition (2.8) we have
π((π₯π+1(π), (π₯π+2(π)) β€ β(π((π₯π(π), (π₯π+1(π)))
Similarly
π((π₯πβ1(π), (π₯π(π)) β€ β(π((π₯πβ2(π), (π₯πβ1(π))
Hence π((π₯π(π), (π₯π+1(π)) β€ β(π((π₯πβ1(π), (π₯π(π))) β€
β2π((π₯
πβ2(π), (π₯πβ1(π))
On continuing this process
π((π₯π(π), (π₯π+1(π)) β€ βπ(π((π₯0(π), (π₯1(π)))
So for π > π
π((π₯π(π), (π₯π(π)) β€ (βπ+ βπ+1+ βπ+2β¦ β¦ β¦ β¦ +
βπβ1) (π((π₯
0(π), (π₯1(π)))
β€ βπ
1ββ(π(π₯0(π), π₯1(π)))
Let 0 βͺ π be given. Choose a natural number π such that
βπ
1ββ(π(π₯0(π), π₯1(π))) βͺ π for every π β₯ π. Thus
π((π₯π(π), (π₯π(π)) β€ βπ
1ββ(π(π₯0(π), π₯1(π))) βͺ π for
every π > π β₯ π.
Therefore the sequence {π₯π(π)} is a Cauchy sequence in
πΊ Γ π. Since (π, π) is complete, there exists π§(π) β πΊ Γ π
such that π₯π(π) β π§(π). Choose a natural number π1 such
that
Hence, we have
π(π§(π), ππ§(π)) β€ π(π§(π), π₯π+1(π)) +
π(π₯π+1(π), ππ§(π))
= π(π§(π), π₯π+1(π)) + π(ππ(π₯π(π), πππ§(π))
β€ π(π§(π), π₯π+1(π)) +
β (π(π₯π(π), π§(π)), [π(π₯π(π), ππ(π₯π(π)), π§(π))],
[π(π₯π(π), πππ§(π)) + π(ππ(π₯π(π), π§(π))])
β€ π(π§(π), π₯π+1(π)) +
β (π(π₯π(π), π§(π)), [π(π§(π), π₯π+1(π)) +
π(π§(π), πππ§(π))],
[π(π₯π(π), πππ§(π)) + π(π₯π+1(π), π§(π))])
Taking π ββ we have
π(π§(π), πππ§(π)) β€ 0 + β (0, [0 +
π(π§(π), πππ§(π))], [π(π§(π), πππ§(π)) + 0])
π(π§(π), πππ§(π)) β€
β (0, π(π§(π), πππ§(π)), π(π§(π), πππ§(π)))
π(π§(π), πππ§(π)) β€ 0
Thusβ(π(π§(π), πππ§(π)) β π. But π(π§(π), πππ§(π)) β π.
Therefore π(π§(π), πππ§(π)) = 0
and so πππ§(π) = π§(π).
Similarly
π(πππ§(π)), π§(π)) β€ π(πππ§(π)), π₯
π+2(π)) +
π(π₯π+2(π), π§(π))
= π(πππ§(π)), πππ₯
π+1(π)) + π(π₯π+2(π), π§(π))
β€ β (π(π§(π), π₯π+1(π)), [π (π§(π), π(π§(π)))
+ π(π₯π+1(π), πππ₯π+1(π))],
[π(π§(π), πππ₯
π+1(π)) + π(πππ§(π)), π₯π+1(π))] +
π(π₯π+2(π), π§(π))
β€ β (π(π§(π), π₯π+1(π)), [π(π§(π), πππ§(π))) +
π(π₯π+1(π), π₯π+1(π))],
[π(π§(π), π₯π+2(π)) + π(π(π§(π), π₯π+1(π))] +
π(π₯π+2(π), π§(π))
Taking π β β we have
π(πππ§(π)), π§(π)) β€
β (π(π§(π), π§(π)), [π(π§(π), π§(π)), [π(π§(π), πππ§(π))) +
π(π§(π), π§(π))],
[π(π§(π), π§(π)) + π(πππ§(π)), π§(π))] + π(π§(π), π§(π))
π(πππ§(π)), π§(π))
β€ β (0, [π(π§(π), πππ§(π))) + 0] , [0
+ π(πππ§(π)), π§(π))] + 0
π(πππ§(π)), π§(π))
β€ β (0, π(π§(π), πππ§(π))) , π(πππ§(π)), π§(π)))
π(πππ§(π)), π§(π)) β€ 0
Thus β (π(πππ§(π)), π§(π))) β π
. But π(πππ§(π)), π§(π)) β
π.
Therefore π(πππ§(π)), π§(π)) = 0 and so ππ(π§(π)) = π§(π).
Hence πππ§(π) = π§(π) = ππ(π§(π)).
π(ππ₯(π), ππ(π¦(π)))β€
β (π(ππ₯(π), ππ¦(π)), π (ππ₯(π), ππ(π₯(π))) +
π (ππ₯(π), ππ(π¦(π))),
π (ππ¦(π), ππ(π¦(π))) +
π (ππ₯(π), ππ(π₯(π)))) β¦ β¦ β¦ β¦ (3.3. (π))
For all π₯, π¦ β π,π β πΊ, then π has a unique fixed point in π. Moreover if (π, π) is a Banach pair, then π and π have unique fixed point in π.
Proof: Let π₯0(π) β πΊ Γ π be arbitrary. Define a sequence
{π₯π}
π(ππ₯π(π), ππ₯π+1(π)) = π(ππ(π₯πβ1(π)), ππ(π₯π(π)))
β€ β (π(ππ₯πβ1(π), ππ₯π(π)), π(ππ₯πβ1(π), ππ(π₯πβ1(π)))
+π (ππ₯π(π), ππ(π₯π(π))) , π(ππ₯πβ1(π), ππ(π₯π(π)))
+π (ππ₯πβ1(π), ππ(π₯πβ1(π))) , π (ππ₯π(π), ππ(π₯π(π)))
+π (ππ₯πβ1(π), ππ(π₯πβ1(π))))
= β (π(ππ₯πβ1(π), ππ₯π(π)), π(ππ₯πβ1(π), ππ₯π(π)) +
π(ππ₯π(π), ππ₯π+1(π)),
π(ππ₯πβ1(π), ππ₯π+1(π)) + π(ππ₯πβ1(π), ππ₯π(π)),
π(ππ₯π(π), ππ₯π(π)) + π(ππ₯πβ1(π), ππ₯π(π)))
= β (π(ππ₯πβ1(π), ππ₯π(π)), π(ππ₯πβ1(π), ππ₯π(π)) +
π(ππ₯π(π), ππ₯π+1(π)),
π(ππ₯πβ1(π), ππ₯π(π)) + π(ππ₯πβ1(π), ππ₯π(π)),
π(ππ₯π(π), ππ₯π+1(π)) + π(ππ₯πβ1(π), ππ₯π(π)))
Hence we get,
π(ππ₯π(π), ππ₯π+1(π)) β€ βπ(ππ₯πβ1(π), ππ₯π(π))
Similarly we can show that,
π(ππ₯π(π), ππ₯π+1(π)) β€ βπ(ππ₯πβ2(π), ππ₯πβ1(π))
In general we can write,
π(ππ₯π(π), ππ₯π+1(π)) β€ βππ(ππ₯0(π), ππ₯1(π))
So for π > π
π(ππ₯π(π), ππ₯π(π)) β€ (βπ+ βπ+1+ βπ+2+ β― β¦ β¦ β¦ +
βπβ1)π(ππ₯
0(π), ππ₯1(π))
β€ βπ
1ββ(π(ππ₯0(π), ππ₯1(π))) βͺ π for every π β₯ π.
Thus π(ππ₯π(π), ππ₯π(π)) β€ βπ
1ββ(π(ππ₯0(π), ππ₯1(π))) βͺ
π π > π β₯ π.
Therefore the sequence {π₯π(π)} is a Cauchy sequence in
πΊ Γ π.
Since (π, π) is complete, there exists π’(π) β πΊ Γ π such that
ππ₯π(π) β ππ’(π).
Since π is subsequently convergent, {π₯π(π)} is such that
lim
πββπ₯π(π) = π§(π).
As π is continuous lim
πββππ₯π(π) = ππ§(π).
By uniqueness of limit π§(π) = ππ’(π) Since π is continuous lim
πββππ₯π(π) = ππ§(π).
Again as π is continuous lim
πββπππ₯π(π) = πππ§(π)
Therefore lim
πββππ₯π+1(π) = πππ§(π)
Choose a natural number π1 such that for every π β₯ π
π(ππ§(π), ππ₯π+1(π)) =βͺ π
2
And π(ππ₯π(π), ππ₯π+1(π)) βͺ π
2
Hence for π β₯ π1 we have
π (ππ(π§(π)), ππ§(π)) β€ π (ππ(π§(π)), ππ₯π+1(π)) +
π(ππ₯π+1(π), ππ§(π))
= π(ππ(π§(π)), ππ(π₯π(π))) + π(ππ₯π+1(π), ππ§(π))
β€ β (π(ππ§(π), ππ₯π(π)), π (ππ§(π), ππ(π§(π))) +
π(ππ₯π(π), ππ(π₯π(π)))
π(ππ§(π), ππ(π₯π(π))) +
π (ππ§(π), ππ(π§(π))) , π(ππ₯π(π), ππ(π§(π)))
+π (ππ§(π), ππ(π§(π))) + π(ππ₯π+1(π), ππ§(π))
= β (π(ππ§(π), ππ₯π(π)), π (ππ§(π), ππ(π§(π))) +
π(ππ₯π(π), ππ(π₯π+1(π)),
π(ππ§(π), ππ₯π+1(π)) +
π (ππ§(π), ππ(π§(π))) , π(ππ₯π(π), ππ(π§(π))) +
π (ππ§(π), ππ(π§(π))) + π(ππ₯π+1(π), ππ§(π))
π (ππ§(π), ππ(π§(π))) β€
π(ππ§(π), ππ₯π+1(π)), π(ππ₯π(π), ππ(π₯π+1(π))
βͺπ
2+ π
2= π for every π β₯ π1
Thus π (ππ§(π), ππ(π§(π))) βͺ π
π for all π β₯ 1.
So π
πβ π (ππ§(π), ππ(π§(π))) β π for all π β₯ 1.
Since ππ β 0 as π β β, and π is closed
βπ (ππ§(π), ππ(π§(π))) β π. But π (ππ§(π), ππ(π§(π))) β π.
Therefore π (ππ§(π), ππ(π§(π))) = 0 And so ππ§(π) = ππ(π§(π)).
As π is injective π§(π) = π(π§(π)). Thus π§(π) is the fixed point of π.
Uniqueness: If π’(π) is another fixed point of π, then
π’(π) = π(π’(π)).
π(ππ’(π), ππ§(π)) = π(ππ(π’(π)), ππ(π§(π)))
β€ β (π(ππ’(π), ππ§(π)), π(ππ’(π), ππ(π’(π)) + π (ππ§(π)ππ(π§(π))),
π(ππ’(π), ππ(π§(π)) + π (ππ’(π), ππ(π’(π))),
π(ππ§(π), ππ(π’(π)) + π(ππ’(π), ππ(π’(π)))= β (π(ππ’(π), ππ§(π)), π(ππ’(π), ππ’(π)) + π(ππ§(π), ππ§(π)),
π(ππ’(π), ππ§(π)) + π(ππ’(π), ππ’(π)),
π(ππ§(π), ππ’(π)) + π(ππ’(π), ππ’(π))
= βπ(ππ’(π), ππ§(π)) as β < 1, a contraction. Hence π(ππ’(π), ππ§(π)) = 0 which implies ππ’(π) = ππ§(π).
As π is injective, π’(π) = π§(π) is the unique fixed point of
π.
As (π, π) is a Banach pair, π and π commutes at fixed point of π which implies that πππ§(π) = πππ§(π) i.e. ππ§(π) = πππ§(π) which implies that ππ§(π) is another fixed point of
π.
By uniqueness of fixed point of π,π§(π) = ππ§(π).
Hence π§(π) = ππ§(π) = ππ§(π) is the unique fixed point of
π and π in π.
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