Volume 2014, Article ID ama0133, 9 pages ISSN 2307-7743
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A UNIQUE COMMON FIXED POINT RESULT IN CONE METRIC SPACE UNDER GENERALIZED
ALTERING DISTANCE FUNCTIONS
TANMOY SOM AND LOKESH KUMAR
Abstract. In the present paper we establish a unique common fixed point result for two-self mappings of a cone metric space involving two generalised altering distance functions. The result obtained generalizes the fixed point result of Choudhury (2005) from metric space to cone metric space.
1. Introduction
On the theory of fixed points for mappings on cone valued metric space, we have some literature since last four decades. Recently the concept of altering distance function is used by Khan et al (1984), Sas-try et al. (1998, 1999), Choudhury (2005) etc. as control function, which has given a new direction to the fixed point theory on metric spaces. In the present work we use this control function on cone metric spaces and obtain a common fixed point result. Before going to our main result, we give here some related preliminary definitions and con-cepts.
LetE be a real Banach space and θ is the zero of the Banach spaceE. LetP be a subset ofE. P is called a cone if
(1) P is closed, non-empty and P 6={θ}
(2) ax+by ∈P for allx, y ∈P and non-negative real numbersa, b
(3) PT
(−P) ={θ}
For a given cone P we can define a partial ordering ≤ with respect to
P by x ≤ y if and only if y−x ∈P. x < y will stand for x ≤ y and
x 6=y, while x << y will stand for y−x∈ intP; where intP denotes the interior ofP. x≤yis same asy≥xandx << yis same asy >> x.
2010Mathematics Subject Classification. 47H10, 54H25.
Key words and phrases. altering distance function, common fixed points.
c
A cone P is called normal if there is a number K > 0 such that for allx, y ∈E,
θ≤x≤y implies kxk ≤Kkyk.
The least positive number satisfying the above inequality is called the normal constant of P. The cone is called regular if every increasing and bounded above sequence xn in E is convergent. Equivalently the cone
P is regular if and only if every decreasing and bounded below sequence is convergent.
Definition 1.1: Let X be a non-empty set. Suppose the mapping
d:X×X →E satisfies :
(1) θ≤d(x, y) for all x, y ∈X and d(x, y) = θ if and only if x=y
(2) d(x, y) =d(y, x) for all x, y ∈X.
(3) d(x, y)≤d(x, z) +d(z, y) for allx, y, z ∈X.
Then d is called a cone metric on X and (X, d) is called a cone metric space.
Definition 1.2: Let (X, d) be a cone metric space, xn a sequence in
X and x∈X. For everyc∈E with θ << c; we say that xn is:
(1) a Cauchy sequence if there is a natural numberN such that for alln, m > N; d(xn, xm)<< c
(2) convergent toxif there is a natural numberN such that for all
n > N;d(xn, x)<< c for some x∈X.
Definition 1.3 (X, d) is called a complete cone metric space if every Cauchy sequence in X is convergent.
2. Some basic Results
Lemma 2.1 ([4]). Let (X, d) be a cone metric space, P be a normal cone with normal constant K. Let {xn} be a sequence in X. Then {xn} converges to x if and only if d(xn, x)−→θ as n−→ ∞.
Lemma 2.2([4]). Let (X, d) be a cone metric space. P be a normal cone with normal constantK. Let{xn}be a sequence inX. Ifxn−→x as well as xn−→y then x=y. That is the limit of{xn} is unique.
Lemma 2.4([4]). Let (X, d) be a cone metric space. P be a normal cone with normal constantK. Let{xn}be a sequence inX. Then{xn} is a Cauchy sequence if and only if d(xn, xm)−→θ as (n, m−→ ∞).
Lemma 2.5: Let E be a real Banach Space and P be a cone. Then
θ /∈intP.
Proof : Let 0 ∈ intP. Then there exist some c(6= θ) ∈ P such that
c << θ.
i.e., θ−c∈P
i.e., c∈ −P, So c∈P T
(−P) i.e., c=θ, a contradiction.
Lemma 2.6: Let E be a real Banach space and P be a cone then P
must be equal to the set {x∈E :θ≤x}, where the partial order ≤is with respect to the cone P.
Proof : Letx∈P then x−θ ∈P. This implies that θ ≤x. If θ≤x, then x−θ ∈P, i.e., x∈P. So x∈P if and only if θ ≤x.
Case 1: IfP =Ethen (−P) = −E =E =P. SoP T
(−P) =P 6={θ}, a contradiction. So P must be a proper subset of E.
Case 2: IfP ={x∈E :θ ≤x≤t}for somet≥θ belonging to E. i.e.,
P has a upper bound. Then for some real number y such that y >1, We have t(y−1)∈P and ty ∈P
i.e., ty−t∈P , becauset ∈P and 0< y−1. i.e., t ≤ty
Now (ty−t) = t(y−1)6=θ, because t6=θ and y−16=θ. i.e., t 6=ty
i.e., t 6=ty and t≤ty.
i.e., t << ty and ty∈P. This is again a contradiction.
SoP must be a set of type {x∈E :θ ≤x}, where the partial order≤ is with respect to the coneP.
3. Main Results
Definition 3.1: A continuous function ψ : P → P is said to be an altering distance function if it satisfies :
(1) ψ is monotone increasing in the sence of norm, i.e. kxk <kyk implieskψ(x)k<kψ(y)k and
(2) ψ(x) = θ if and only if x=θ.
Definition 3.2: A continuous function ψ : P3 → P is said to be a generalized altering distance function if
(1) ψ is monotone increasing in all the three variables in the sence of norm, and
(2) ψ(x, y, z) = θ if and only if x=y=z =θ.
Theorem 3.1: Let P be the normal cone. Let (X, d) be a complete cone metric space and S and T be two self- mappings such that the following inequality is satisfied:
kϕ1(d(Sx, T y))k ≤ kψ1(d(x, y), d(x, Sx), d(y, T y))k
− kψ2(d(x, y), d(x, Sx), d(y, T y))k
(1)
where ψ1 and ψ2 are generalized altering distance functions and
ϕi(x) =ψi(x, x, x), where i= 1,2.
Then S and T have a unique common fixed point.
Proof: Let x0 ∈X be an arbitrary point. Define
(2) x2n+1 =Sx2n for n = 0,1,2,3· · ·
and
(3) x2n+2 =T x2n+1, for n= 0,1,2,3· · ·
Let an=d(xn, xn+1). Putting x=x2n and y=x2n+1 in (1),we get for
alln = 0, 1, 2 · · ·
kϕ1(d(Sx2n, T x2n+1))k
≤ kψ1(d(x2n, x2n+1), d(x2n, Sx2n), d(x2n+1, T x2n+1))k
− kψ2(d(x2n, x2n+1), d(x2n, Sx2n), d(x2n+1, T x2n+1))k
i.e.,
kϕ1(d(x2n+1, x2n+2))k
≤ kψ1(d(x2n, x2n+1), d(x2n, x2n+1), d(x2n+1, x2n+2))k
− kψ2(d(x2n, x2n+1), d(x2n, x2n+1), d(x2n+1, x2n+2))k
kϕ1(a2n+1)k
≤ kψ1(a2n, a2n, a2n+1)k − kψ2(a2n, a2n, a2n+1)k
(4)
Ifka2nk<ka2n+1k, then 0<kψ2(a2n, a2n, a2n+1)k. This implies
kϕ1(a2n+1)k<kψ1(a2n, a2n, a2n+1)k
<kψ1(a2n+1, a2n+1, a2n+1)k
i.e.,
kϕ1(a2n+1)k<kϕ1(a2n+1)k
This is a contradiction. So for n = 0, 1, 2 · · ·
(5) ka2n+1k ≤ ka2nk
Putting x=x2n and y=x2n−1 in (1) we obtain
kϕ1(a2n)k
≤ kψ1(a2n−1, a2n−1, a2n)k − kψ2(a2n−1, a2n−1, a2n)k
(6)
Similarly we can show that,
(7) ka2n+2k ≤ ka2n+1k for all n= 0,1,2,3· · ·
From (5) and (7), we obtain,
(8) kan+1k ≤ kank for all n= 0,1,2,3· · ·
Then from (4) and (6) we obtain, for all n = 0, 1, 2· · ·
kϕ1(an+1)k ≤ kϕ1(an)k − kϕ2(an+1k)
i.e.,
kϕ2(an+1)k ≤ kϕ1(an)k − kϕ1(an+1)k
Summing up from 0 to ∞ we obtain
P
||ϕ2(an+1)k ≤ kϕ1(ao)k < ∞, i.e.,
P
||ϕ2(an+1)k is a real
conver-gent series. So it’s nth term must goes to zero as n tends to infinity. i.e., kϕ2(an)k →0 as n → ∞
kank −→a asn −→ ∞, for some real number a. Now we will prove thatan−→θ. We have
(9) kϕ2(an)k →0 as n→ ∞
i.e.,
limn→∞kϕ2(an)k= 0
⇒ klimn→∞ϕ2(an)k= 0
⇒ kϕ2(limn→∞an)k= 0 ⇒ϕ2(limn→∞an) =θ ⇒limn→∞an=θ
i.e.,
(10) limn→∞d(xn, xn+1) = θ
We now prove that {xn} is a Cauchy sequence. In the view of (10) it is sufficient to prove that {x2r} ⊂ {xn}is Cauchy sequence. If {x2r} is not a Cauchy sequence, then given
∈intP such that kk>0 we can find monotone increasing sequences of natural numbers {2m(k)} and {2n(k)} such that, for n(k)> m(k),
kd(x2m(k), x2n(k))k>kk and
kd(x2m(k), x2n(k)−1)k<kk
(11)
Then
kk<kd(x2m(k), x2n(k))k
≤ kd(x2m(k), x2n(k)−1) +d(x2n(k)−1, x2n(k))k
≤ kd(x2m(k), x2n(k)−1)k+kd(x2n(k)−1, x2n(k))k
<kk+kd(x2n(k)−1, x2n(k))k
Taking k→ ∞ in the above inequality we obtain
(12) limk→∞kd(x2m(k), x2n(k))k=kk
Now, for all k = 1, 2,· · ·
kd(x2n(k)+1, x2m(k))k
≤ kd(x2n(k)+1, x2n(k)) +d(x2n(k), x2m(k))k
i.e.,
kd(x2n(k)+1, x2m(k))k
≤ kd(x2n(k)+1, x2n(k))k+kd(x2n(k), x2m(k))k
and
kd(x2n(k), x2m(k))k
≤ kd(x2n(k), x2n(k)+1) +d(x2n(k)+1, x2m(k))k
i.e.,
kd(x2n(k), x2m(k))k
≤ kd(x2n(k), x2n(k)+1)k+kd(x2n(k)+1, x2m(k))k
(14)
Taking k → ∞ in the above inequalities and using (10) and (12) we obtain
limk→∞kd(x2n(k)+1, x2m(k))k ≤ kk
and
kk ≤limk→∞kd(x2n(k)+1, x2m(k))k
i.e.,
(15) limk→∞kd(x2n(k)+1, x2m(k))k=kk
Now for all k = 1, 2,· · ·
kd(x2n(k), x2m(k)−1)k
≤ kd(x2n(k), x2m(k)) +d(x2m(k), x2m(k)−1)k
≤ kd(x2n(k), x2m(k))k+kd(x2m(k), x2m(k)−1)k
and
kd(x2n(k), x2m(k))k
≤ kd(x2n(k), x2m(k)−1) +d(x2m(k)−1, x2m(k))k
≤ |d(x2n(k), x2m(k)−1)k+kd(x2m(k)−1, x2m(k))k
Taking k → ∞ in the above inequalities and using (10) and (15) we obtain
(16) limk→∞kd(x2n(k), x2m(k)−1))k=kk
Now puttingx=x2n(k)and y=x2m(k)−1 in (1) we obtain for allk ∈N
kϕ1(d(x2n(k)+1, x2m(k)))k
≤ kψ1(d(x2n(k), x2m(k)−1), d(x2n(k), x2n(k)+1),
(x2m(k)−1, x2m(k)))k
− kψ2(d(x2n(k), x2m(k)−1), d(x2n(k), x2n(k)+1),
d(x2m(k)+1, x2m(k)))k
(17) kϕ1()k ≤ kψ1(,0,0)k − kψ2(,0,0)k<kϕ1()k
This is due to the fact that ψ1 and ψ2 are monotone increasing in all
the three variables in the sense of norm and ψ(x, y, z) =θ if and only if x = y = z = θ So if we assume that {x2r}is not Cauchy sequence then we finally achieve a contradiction. That means {x2r}is a Cauchy sequence hence by (10), {xn} is also a Cauchy sequence.
As (X, d) is a complete cone metric space {xn} will converge to some point z (say) belonging to X.
Putting x=x2n and y=z in (1) we obtain for alln ∈N
kϕ1(d(x2n+1, T z))k
≤ kψ1(d(x2n, z), d(x2n, x2n+1), d(z, T z))k
− kψ2(d(x2n, z), d(x2n, x2n+1), d(z, T z))k
Taking n → ∞ in the above inequality, we obtain by using (9) and
xn→z as n→ ∞ with continuity of ψ1 and ψ2, we get,
kϕ1(d(z, T z))k ≤ kψ1(θ, θ, d(z, T z))k − kψ2(θ, θ, d(z, T z))k
If d(z, T z)6=θ then 0<kψ2(d(θ, θ, d(z, T z))k
So
kϕ1(d(z, T z))k<kψ1(θ, θ, d(z, T z))k
<kψ1(d(z, T z), d(z, T z), d(z, T z))k
=kϕ1(d(z, T z))k
i.e., kϕ1(d(z, T z))k<kϕ1(d(z, T z))k.
This is a contradiction.
i.e., (d(z, T z)) must be equal toθ. i.e., T z =z.
Similarly we can prove that Sz =z.
i.e., z is a common fixed point of S and T.
Uniqueness : Let w(6= z) be another common fixed point of S and T. Then d(z, w)6=θ and
kϕ1(d(Sz, T w))k=kϕ1(d(z, w))k
≤ kψ1(d(z, w), θ, θ)k − kψ2(d(z, w), θ, θ)k <kϕ1(d(z, w))k
Corollary 3.1: Let (X, d) is a complete cone metric space and S, T :
X →X which satisfy :
3kd(Sx, T y)k ≤ kd(x, y) +d(x, Sx) +d(y, T y)k −k kmax{d(x, y), d(x, Sx), d(y, T y)}k (18)
wherekis a positive real number, thenSandT have a unique common fixed point.
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