Abstract In this paper we establish certain multivalued and singlevalued fixed pointresults under the assumptions of certain almost contractive type inequalities. Ourresults are derived in metric spaces with a partial ordering. We have given three illus-trative examples. The results proved here extend several existing results. Two of ourexamples show that the improvements are actual.
Keywords Partial ordering · Almost contraction · Fixed point
Let (X, d) be a metric space. We denote the class of nonempty and bounded subsets ofX by B(X). For A, B ∈ B(X), functions D(A, B) and δ(A, B) are defined as follows:
D(A, B) = inf {d(a, b) : a ∈ A, b ∈ B},δ(A, B) = sup {d(a, b) : a ∈ A, b ∈ B}.
B. S. ChoudhuryDepartment of Mathematics, Bengal Engineering and Science University,Shibpur, Howrah 711103, West Bengal, Indiae-mail: [email protected]; [email protected]
N. Metiya (B)Department of Mathematics, Bengal Institute of Technology,Kolkata 700150, West Bengal, Indiae-mail: [email protected]
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If A = {a}, then we write D(A, B) = D(a, B) and δ(A, B) = δ(a, B). Also inaddition, if B = {b}, then D(A, B) = d(a, b) and δ(A, B) = d(a, b). Obviously,D(A, B) ≤ δ(A, B). For all A, B,C ∈ B(X), the definition of δ(A, B) yields thefollowing:
δ(A, B) = δ(B, A),
δ(A, B) ≤ δ(A,C)+ δ(C, B),
δ(A, B) = 0 iff A = B = {a},δ(A, A) = diam A. [1, 2]
Fixed point theory of multivalued functions is a vast chapter of functional analysis.In particular, the function δ(A, B) has been used in many works in this area. Some ofthese works are noted in [1–6].
We will use the following relation between two nonempty subsets of a partiallyordered set.
Definition 1.1 [4] Let A and B be two nonempty subsets of a partially ordered set(X,�). The relation between A and B is denoted and defined as follows:
(i) A ≺1 B, if for every a ∈ A there exists b ∈ B such that a � b,(ii) A ≺2 B, if for every b ∈ B there exists a ∈ A such that a � b,
(iii) A ≺3 B, if A ≺1 B and A ≺2 B.
In recent times, fixed point theory has developed rapidly in partially ordered metricspaces. An early result in this direction was established by Turinici in ordered metriz-able uniform spaces [7]. Application of fixed point results in partially ordered metricspaces were made subsequently, for example, by Ran and Reurings [8] to solvingmatrix equations and by Nieto and Rodriguez-López [9] to obtain solutions of certainpartial differential equations with periodic boundary conditions. Some more recentreferences in which new fixed point results have been obtained in such spaces arenoted in [10–13].
The concept of almost contractions were introduced by Berinde [14,15]. It wasshown in [14] that any strict contraction, the Kannan [16] and Zamfirescu [17] map-pings, as well as a large class of quasi-contractions, are all almost contractions. Almostcontractions and its generalizations were further considered in several works like [18–28]. Recently, in their paper, Ciric et al. [29], proved some fixed point results in orderedmetric spaces using almost generalized contractive condition, which is given in thefollowing definition.
Definition 1.2 [29] Let f and g be two self maps on a metric space (X, d). They aresaid to satisfy almost generalized contractive condition if there exists δ ∈ (0, 1) andL ≥ 0 such that
d( f x, gy) ≤ δ max
{d(x, y), d(x, f x), d(y, gy),
d(x, gy)+ d(y, f x)
2
}
+L min {d(x, f x), d(y, gy), d(x, gy), d(y, f x)}, for all x, y ∈ X.
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Ann Univ Ferrara (2012) 58:21–36 23
The purpose of this paper is to establish the existence of fixed points of multi-valued and singlevalued mappings in partially ordered metric spaces. The mappingsare assumed to satisfy certain almost contractive type inequalities. Further we haveestablished that in the corresponding singlevalued cases a partial order condition ofthe metric space can be omitted if the function is continuous. We have concluded ourpaper with three illustrative examples.
2 Main results
Theorem 2.1 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T : X −→ B(X) be amultivalued mapping such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that {x0} ≺1 T x0,(ii) for x, y ∈ X, x � y implies T x ≺1 T y,
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) δ(T x, T y)
≤ ψ
(max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})
+L min {D(x, T x), D(y, T y), D(x, T y), D(y, T x)},
for all comparable x, y ∈ X, where L ≥ 0 and ψ : [0,∞) −→ [0,∞) isa nondecreasing and upper-semi continuous function with ψ(t) < t for eacht > 0.
Then T has a fixed point.
Proof By the assumption (i), there exists x1 ∈ T x0 such that x0 � x1. By the assump-tion (ii), T x0 ≺1 T x1. Then there exists x2 ∈ T x1 such that x1 � x2. Continuing thisprocess we construct a monotone increasing sequence {xn} in X such that xn+1 ∈ T xn ,for all n ≥ 0. Thus we have x0 � x1 � x2 � x3 � · · · � xn � xn+1 � · · ·.
If there exists a positive integer N such that xN = xN+1, then xN is a fixed pointof T. Hence we shall assume that xn = xn+1, for all n ≥ 0.
Using the condition (iv), we have for all n ≥ 0,
d(xn+1, xn+2) ≤ δ(T xn, T xn+1)
≤ ψ
(max
{d(xn, xn+1), D(xn, T xn), D(xn+1, T xn+1),
D(xn, T xn+1)+ D(xn+1, T xn)
2
})
+ L min {D(xn, T xn), D(xn+1, T xn+1), D(xn, T xn+1), D(xn+1, T xn)}
Suppose that d(xn, xn+1) ≤ d(xn+1, xn+2), for some positive integer n.Then from (2.1), we have
d(xn+1, xn+2) ≤ ψ(d(xn+1, xn+2)),
which implies that d(xn+1, xn+2) = 0, or that xn+1 = xn+2, contradicting our assump-tion that xn = xn+1, for each n.
Therefore, d(xn+1, xn+2) < d(xn, xn+1), for all n ≥ 0 and {d(xn, xn+1)} is amonotone decreasing sequence of nonnegative real numbers. Hence there exists ar ≥ 0 such that
d(xn, xn+1) −→ r as n −→ ∞. (2.2)
Now, from (2.1), we have
d(xn+1, xn+2) ≤ ψ(d(xn, xn+1)).
Taking n −→ ∞ in the above inequality and using the properties of ψ , we have
r ≤ limn→∞ supψ(d(xn, xn+1)) ≤ ψ(r),
which is a contradiction unless r = 0. Thus we have
limn→∞d(xn, xn+1) = 0. (2.3)
Next we show that {xn} is a Cauchy sequence. If otherwise, there exists an ε > 0 forwhich we can find two sequences of positive integers {m(k)} and {n(k)} such that forall positive integers k, n(k) > m(k) > k and d(xm(k), xn(k)) ≥ ε.
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Ann Univ Ferrara (2012) 58:21–36 25
Assuming that n(k) is the smallest such positive integer, we get
+ L min{d(xm(k), xm(k)+1), d(xn(k), xn(k)+1), d(xm(k), xn(k)+1),
d(xn(k), xm(k)+1)}.
Letting k −→ ∞ in the above inequality, using (2.3), (2.4), (2.5), (2.6) and (2.7) andusing the properties of ψ , we have
ε ≤ ψ(ε) < ε,
which is a contradiction.Hence {xn} is a Cauchy sequence. From the completeness of X, there exists a z ∈ X
such that
xn −→ z as n −→ ∞. (2.8)
By the assumption (iii), xn � z, for all n.Then by the condition (iv), we have
δ(xn+1, T z) ≤ δ(T xn, T z)
≤ ψ
(max
{d(xn, z), D(xn, T xn), D(z, T z),
D(xn, T z)+ D(z, T xn)
2
})
+ L min {D(xn, T xn), D(z, T z), D(xn, T z), D(z, T xn)}≤ ψ
(max
{d(xn, z), d(xn, xn+1), D(z, T z),
D(xn, T z)+ d(z, xn+1)
2
})
+ L min {d(xn, xn+1), D(z, T z), D(xn, T z), d(z, xn+1)} .
Taking the limit as n −→ ∞ in the above inequality, using (2.3) and (2.8) and theproperties of ψ , we have
δ(z, T z) ≤ ψ(D(z, T z)) ≤ ψ(δ(z, T z)),
which implies δ(z, T z) = 0, or that {z} = T z. Moreover, z is a fixed point of T. �
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Example 2.1 Let X = {(0, 0), (− 12 ,− 1
2 ), (0,−1)} be a subset of R2 with the order� defined as: for (x1, y1), (x2, y2) ∈ X , (x1, y1) � (x2, y2) if and only if x1 ≤ x2,y1 ≤ y2. Let d : X × X −→ R be given as
d(x, y) = max {|x1 − x2|, |y1 − y2|}, for x = (x1, y1), y = (x2, y2) ∈ X.
Then (X, d) is a complete metric space with the required properties of Theorem 2.1.
Let T : X −→ B(X) be defined as follows:
T x =⎧⎨⎩
{(0, 0)}, if x = (0, 0),{(0, 0), (− 1
2 ,− 12 )}, if x = (0,−1),
{(0, 0)}, if x = (− 12 ,− 1
2 ).
Then T has the properties mentioned in Theorem 2.1.Let ψ : [0,∞) −→ [0,∞) be defined as follows:
ψ(t) ={
t2
2 , if t ≤ 1,t2 , if t > 1.
Then ψ has the properties mentioned in Theorem 2.1.Without loss of generality, we assume that x � y and discuss the following cases.
(i) If x = (0,−1) and y = (0, 0), then
δ(T x, T y) = 1
2,
max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})= 1 and
min {D(x, T x), D(y, T y), D(x, T y), D(y, T x)} = 0.
(ii) If x = (− 12 ,− 1
2 ) and y = (0, 0), then
δ(T x, T y) = 0,
max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})= 1
2and
min{D(x, T x), D(y, T y), D(x, T y), D(y, T x)} = 0.
(iii) If x = (0, 0) and y = (0, 0), then
δ(T x, T y) = 0,
max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})= 0 and
min {D(x, T x), D(y, T y), D(x, T y), D(y, T x)} = 0.
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(iv) If x = (− 12 ,− 1
2 ) and y = (− 12 ,− 1
2 ), then
δ(T x, T y) = 0,
max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})= 1
2and
min {D(x, T x), D(y, T y), D(x, T y), D(y, T x)} = 1
2.
(v) If x = (0,−1) and y = (0,−1), then
δ(T x, T y) = 1
2,
max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})= 1
2and
min {D(x, T x), D(y, T y), D(x, T y), D(y, T x)} = 1
2.
Clearly, for all comparable x, y ∈ X and L = 1,
δ(T x, T y) ≤ ψ
(max
{d(x, y), D(x, T x), D(y, T y),
D(x, T y)+ D(y, T x)
2
})
+L min, {D(x, T x), D(y, T y), D(x, T y), D(y, T x)}.
Hence the conditions of Theorem 2.1 are satisfied and it is seen that (0, 0) is a fixedpoint of T .
Remark 2.1 Taking L = 0 andψ(t) = αt , where 0 < α < 1, in Theorem 2.1, we havecorollary 2 of [4]. The above example does not satisfy the Theorem 2.1, when L = 0and ψ(t) = αt , where 0 < α < 1. Hence Theorem 2.1 is an actual improvement overthe Corollary 2 of [4].
The following corollary is a special case of Theorem 2.1 when T is a singlevaluedmapping.
Corollary 2.2 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T : X −→ X be amapping such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that x0 � T x0,(ii) for x, y ∈ X, x � y implies T x � T y,
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) d(T x, T y) ≤ ψ
(max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})
+L min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)},
for all comparable x, y ∈ X, where L ≥ 0 and ψ : [0,∞) −→ [0,∞) isa nondecreasing and upper-semi continuous function with ψ(t) < t for eacht > 0.
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Ann Univ Ferrara (2012) 58:21–36 29
Then T has a fixed point.
In the following theorem we consider T a singlevalued mapping and replace con-dition (iii) of Theorem 2.1 by requiring T to be continuous.
Theorem 2.3 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T : X −→ X be acontinuous mapping such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that x0 � T x0,(ii) for x, y ∈ X, x � y implies T x � T y,
(iii) d(T x, T y) ≤ ψ
(max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})
+L min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)},
for all comparable x, y ∈ X, where L ≥ 0 and ψ : [0,∞) −→ [0,∞) isa nondecreasing and upper-semi continuous function with ψ(t) < t for eacht > 0.
Then T has a fixed point.
Proof We can treat T as a multivalued mapping in which case T x is a singleton setfor every x ∈ X . Then we consider the same sequence {xn} as in the proof of Theo-rem 2.1. Arguing exactly as in the proof of Theorem 2.1, we have that {xn} is a Cauchysequence and limn→∞ xn = z. Then, the continuity of T implies that
z = limn→∞ xn+1 = lim
n→∞ T xn = T z
and this proves that z is a fixed point of T . �Example 2.2 Let X = {(0, 0), (− 1
2 , 0), (0,−1)} be a subset of R2 with the order �defined as: for (x1, y1), (x2, y2) ∈ X , (x1, y1) � (x2, y2) if and only if x1 ≤ x2,y1 ≤ y2. Let d : X × X −→ R be given as
d(x, y) = max {|x1 − x2|, |y1 − y2|}, for x = (x1, y1), y = (x2, y2) ∈ X.
Then (X, d) is a complete metric space with the required properties of Corollary 2.2and Theorem 2.3.
Let T : X −→ X be defined as follows:
T x =⎧⎨⎩(0, 0), if x = (0, 0),(− 1
2 , 0), if x = (0,−1),(0, 0), if x = (− 1
2 , 0).
Then T has the properties mentioned in Corollary 2.2 and Theorem 2.3.
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30 Ann Univ Ferrara (2012) 58:21–36
Let ψ : [0,∞) −→ [0,∞) be defined as follows:
ψ(t) ={
t2
2 , if t ≤ 1,t2 , if t > 1.
Then ψ has the properties mentioned in Corollary 2.2 and Theorem 2.3.Without loss of generality, we assume that x � y and discuss the following cases.
(i) If x = (0,−1) and y = (0, 0), then
d(T x, T y) = 1
2,
max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})= 1 and
min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} = 0.
(ii) If x = (− 12 , 0) and y = (0, 0), then
d(T x, T y) = 0,
max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})= 1
2, and
min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} = 0.
(iii) If x = (0, 0) and y = (0, 0), then
d(T x, T y) = 0,
max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})= 0 and
min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} = 0.
(iv) If x = (− 12 , 0) and y = (− 1
2 , 0), then
d(T x, T y) = 0,
max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})= 1
2and
min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} = 1
2.
(v) If x = (0,−1) and y = (0,−1), then
d(T x, T y) = 0,
max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})= 1 and
min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} = 1.
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Ann Univ Ferrara (2012) 58:21–36 31
Clearly, for all comparable x, y ∈ X and L ≥ 0,
d(T x, T y) ≤ ψ
(max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})
+ L min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)}.
Hence the conditions of Corollary 2.2 and Theorem 2.3 are satisfied and it is seen that(0, 0) is a fixed point of T .
Remark 2.2 Taking ψ(t) = αt , where 0 < α < 1 in Theorem 2.3 and Corollary 2.2,we have Theorem 2.1 and Theorem 2.2 of [29] respectively. The above example doesnot satisfy the Theorem 2.3 and Corollary 2.2, when ψ(t) = αt , where 0 < α < 1.Hence Theorem 2.3 and Corollary 2.2 are actual improvements over Theorem 2.1 andTheorem 2.2 of [29] respectively.
Example 2.3 Let X = [0, 1] with usual order � be a partially ordered set. Let d :X × X −→ R be given as d(x, y) = |x − y|, for x, y ∈ X . Then (X, d) is a completemetric space with the required properties of Corollary 2.2.
Let T : X −→ X be defined as follows:
T x ={
0, if 0 ≤ x ≤ 12 ,
132 , if 1
2 < x ≤ 1.
Then T has the properties mentioned in Corollary 2.2.Let ψ : [0,∞) −→ [0,∞) be defined as follows:
ψ(t) ={
t2
2 , if t ≤ 1,t2 , if t > 1.
Then ψ has the properties mentioned in Corollary 2.2.It can be verified that for all comparable x, y ∈ X and L ≥ 0,
d(T x, T y) ≤ ψ
(max
{d(x, y), d(x, T x), d(y, T y),
d(x, T y)+ d(y, T x)
2
})
+ L min {d(x, T x), d(y, T y), d(x, T y), d(y, T x)} .
Hence the conditions of Corollary 2.2 are satisfied and it is seen that 0 is a fixed pointof T .
Theorem 2.4 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T, S : X −→ B(X)be two multivalued mappings such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that {x0} ≺1 T x0,(ii) for x, y ∈ X, x � y implies Sy ≺3 T x,
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32 Ann Univ Ferrara (2012) 58:21–36
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) δ(T x, Sy) ≤ αmax
{d(x, y), D(x, T x), D(y, Sy),
D(x, Sy)+ D(y, T x)
2
}
+L min {D(x, T x), D(y, Sy), D(x, Sy), D(y, T x)},
for all comparable x, y ∈ X, where 0 < α < 1 and L ≥ 0.
Then T and S have a common fixed point. Moreover, any fixed point of T is a fixedpoint of S and conversely.
Proof By the assumption (i), there exists x1 ∈ T x0 such that x0 � x1. By the assump-tion (ii), Sx1 ≺3 T x0 which implies Sx1 ≺2 T x0. Then there exists x2 ∈ Sx1 suchthat x2 � x1. Again, by the assumption (ii), Sx1 ≺3 T x2 which implies Sx1 ≺1 T x2.Then there exists x3 ∈ T x2 such that x2 � x3. Continuing in this manner we con-struct a sequence {xn} such that x2n+1 ∈ T x2n and x2n+2 ∈ Sx2n+1 and x2n � x2n+1,x2n+2 � x2n+1, for n = 0, 1, 2, . . .; that is, consecutive terms are comparable.
First we prove any fixed point of T is a fixed point of S and conversely.Now suppose that p is a fixed point of T but not a fixed point S. By the condition
(iv), we have
δ(p, Sp) ≤ δ(T p, Sp)
≤ αmax
{0, 0, D(p, Sp),
1
2[D(p, Sp)+ 0]
}
+L min {0, D(p, Sp), D(p, Sp), 0},
that is, δ(p, Sp) ≤ αD(p, Sp) ≤ αδ(p, Sp), which is a contradiction unlessδ(p, Sp) = 0; or that, {p} = Sp; that is, p is a fixed point of S.
Using a similar argument, we have that any fixed point of S is also a fixed point of T .If there exists a positive integer 2N such that x2N = x2N+1, then x2N is a fixed
point of T and hence a fixed point of S. A similar conclusion holds if x2N+1 = x2N+2,for some N . Therefore, we may assume that xn = xn+1, for all n ≥ 0.
By the condition (iv), we have for all n ≥ 0,
d(x2n+1, x2n+2) ≤ δ(T x2n, Sx2n+1)
≤ αmax
{d(x2n, x2n+1), D(x2n, T x2n), D(x2n+1, Sx2n+1),
D(x2n, Sx2n+1)+ D(x2n+1, T x2n)
2
}
+L min {D(x2n, T x2n), D(x2n+1, Sx2n+1), D(x2n, Sx2n+1), D(x2n+1, T x2n)}≤ αmax
which implies that {xn} is a Cauchy sequence. From the completeness of X, thereexists a z ∈ X such that
xn −→ z as n −→ ∞. (2.12)
By the assumption (iii), xn � z, for all n.Then by the condition (iv), we have
δ(T z, x2n+2) ≤ δ(T z, Sx2n+1)
≤ αmax
{d(z, x2n+1), D(z, T z), D(x2n+1, Sx2n+1),
1
2[D(z, Sx2n+1)+ D(x2n+1, T z)]
}
+ L min {D(z, T z), D(x2n+1, Sx2n+1), D(z, Sx2n+1), D(x2n+1, T z)}≤ αmax
{d(z, x2n+1), D(z, T z), d(x2n+1, x2n+2),
1
2[d(z, x2n+2)+ D(x2n+1, T z)]
}
+L min {D(z, T z), d(x2n+1, x2n+2), d(z, x2n+2), D(x2n+1, T z)} .
Taking the limit as n −→ ∞ in the above inequality, using (2.11) and (2.12), we have
δ(T z, z) ≤ αD(z, T z) ≤ αδ(z, T z),
which is a contradiction unless δ(z, T z) = 0, or that {z} = T z; that is, z is a fixedpoint of T. By what we have already proved, z is a common fixed point of T and S.
�Arguing in the same way, we can establish the fixed point result noted in the fol-
lowing theorem.
Theorem 2.5 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T, S : X −→ B(X)be two multivalued mappings such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that T x0 ≺2 {x0},(ii) for x, y ∈ X, x � y implies T y ≺3 Sx,
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) δ(T x, Sy) ≤ αmax
{d(x, y), D(x, T x), D(y, Sy),
D(x, Sy)+ D(y, T x)
2
}
+L min {D(x, T x), D(y, Sy), D(x, Sy), D(y, T x)},
for all comparable x, y ∈ X, where 0 < α < 1 and L ≥ 0.
Then T and S have a common fixed point. Moreover, any fixed point of T is a fixedpoint of S and conversely.
Proof The proof is similar that of Theorem 2.4.The following theorems are analogous to Theorems 2.4 and 2.5.
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Ann Univ Ferrara (2012) 58:21–36 35
Theorem 2.6 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T, S : X −→ B(X)be two multivalued mappings such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that {x0} ≺1 Sx0,(ii) for x, y ∈ X, x � y implies T y ≺3 Sx,
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) δ(T x, Sy) ≤ αmax
{d(x, y), D(x, T x), D(y, Sy),
D(x, Sy)+ D(y, T x)
2
}
+L min {D(x, T x), D(y, Sy), D(x, Sy), D(y, T x)},
for all comparable x, y ∈ X, where 0 < α < 1 and L ≥ 0.
Then T and S have a common fixed point. Moreover, any fixed point of T is a fixedpoint of S and conversely.
Theorem 2.7 Let (X,�) be a partially ordered set and suppose that there exists ametric d on X such that (X, d) is a complete metric space. Let T, S : X −→ B(X)be two multivalued mappings such that the following conditions are satisfied:
(i) there exists x0 ∈ X such that Sx0 ≺2 {x0},(ii) for x, y ∈ X, x � y implies Sy ≺3 T x,
(iii) if xn −→ x is any sequence in X whose consecutive terms are comparable,then xn � x, for all n,
(iv) δ(T x, Sy) ≤ αmax
{d(x, y), D(x, T x), D(y, Sy),
D(x, Sy)+ D(y, T x)
2
}
+ L min {D(x, T x), D(y, Sy), D(x, Sy), D(y, T x)},
for all comparable x, y ∈ X, where 0 < α < 1 and L ≥ 0.
Then T and S have a common fixed point. Moreover, any fixed point of T is a fixedpoint of S and conversely.
Remark 2.3 Taking L = 0, in Theorems 2.4–2.7, we have Theorems 2–5 of [4] respec-tively. �Acknowledgments The authors gratefully acknowledge the suggestions made by the learned referee. Thework is partially supported by Council of Scientific and Industrial Research, India (Project No. 25(0168)/09/EMR-II). B.S. Choudhury gratefully acknowledges the support.
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