Journal of Advanced Research in
Pure Mathematics
Online ISSN: 1943-2380
Vol. 6, Issue. 4, 2014, pp. 9-23
doi: 10.5373/jarpm.1896.120413
Fixed points of ψ-weak Geraghty contractions
in partially ordered metric spaces
G.V.R. Babu1,∗ , K.K.M. Sarma1 , P.H. Krishna2
1
Department of Mathematics, Andhra University, Visakhapatnam-530 003, India.
Department of Mathematics, Viswanadha Institute of Technology and Management,
Visakhapatnam-531 173, India.
2
Abstract. We introduce ψ-weak generalized Geraghty contractions and prove the
existence of fixed points in partially ordered complete metric spaces, where ψ is an
altering distance function. Our results extend and generalize the results of Gordji,
Ramezani, Cho and Pirbavafa [6], Choudhury and Kundu [4].
Keywords: Fixed points; Partially ordered metric spaces; Weak contraction; Altering distance
function.
Mathematics Subject Classification 2010: 47H10, 54H25.
1
Introduction
Banach contraction principle deals with the existence and uniqueness of fixed points
of contraction mappings in complete metric spaces. This theorem has been extended and
generalized by many researchers. One among those generalizations is due to Geraghty.
In 1973, Geraghty [5] introduced an extension of the contraction in which the contraction
constant was replaced by a function having some specific properties.
Throughout this paper, S denotes the class of functions
β : R+ → [0, 1) satisfying the following: for a sequence {tn } ⊂ R+ = [0, ∞),
β(tn ) → 1 ⇒ tn → 0.
Definition 1.1. [5] Let (X, d) be a metric space. A selfmap f : X → X is said to be a
Geraghty contraction if there exists β ∈ S such that
d(f (x), f (y)) ≤ β(d(x, y))d(x, y)
”(1)”
for all x, y ∈ X.
∗
Correspondence to: G.V.R. Babu, Department of Mathematics, Andhra University, Visakhapatnam-530
003, India. Email: gvr [email protected].
†
Received: 4 December 2013, accepted: 14 May 2014.
http://www.i-asr.com/Journals/jarpm/
9
c
⃝2014
Institute of Advanced Scientific Research
10
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
Here we observe that every contraction is a Geraghty contraction, but its converse
need not be true [3].
Theorem 1.2. [5] Let (X, d) be a complete metric space. Let f : X → X be a
Geraghty contraction. Then for any choice of initial point x0 ∈ X, the iteration {xn }
defined by xn = f (xn−1 ) for n = 1, 2, 3, ... converges to the unique fixed point z of f in
X .
In 2004, Ran and Reurings [12] generalized Banach contraction principle by
introducing a partial order in metric spaces.
Definition 1.3. Let (X, ≼) be a partially ordered set. A mapping f : X → X is said
to be non-decreasing if for any x, y ∈ X, x ≼ y ⇒ f (x) ≼ f (y).
In 2010, Amini-Harindi and Emami [1] extended Theorem 1.2 to partially ordered
metric spaces.
Theorem 1.4. [1] Let (X, ≼) be a partially ordered set and suppose that there exists
a metric d on X such that (X, d) is a complete metric space. Let f : X → X be an
increasing mapping such that there exists x0 ∈ X with x0 ≼ f (x0 ). Suppose that there
exists β ∈ S such that
d(f (x), f (y)) ≤ β(d(x, y))d(x, y) for all x, y ∈ X with x ≽ y.
”(2)”
Assume that either
(i) f is continuous (or)
(ii) X is such that if an increasing sequence {xn } → x in X, then xn ≼ x for all n.
Further, assume that for each x, y ∈ X there exists z ∈ X such that z is comparable
to x and y.
Then f has a unique fixed point in X.
For more works on the existence of fixed points in partially ordered sets, we
refer [7–9], and [13].
The following class of functions, namely, the class of altering distance functions which
we denote by Ψ, were introduced by Khan, Swaleh and Sessa [11] .
Ψ = {ψ : R+ → R+ /ψ is non-decreasing, continuous and ψ(t) = 0 ⇔ t = 0}. Here we
note that any function ψ ∈ Ψ need not be subadditive.
Sastry and Babu [14] continued the study of existence of fixed points by using the
technique altering distances.
In 2012, Gordji, Ramezani, Cho and Pirbavafa [6] proved the following theorem by
using an element ψ ∈ Ψ along with the additional property namely sub-additivity.
i.e., ψ(s + t) ≤ ψ(s) + ψ(t) for all s, t ≥ 0.
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
11
Theorem 1.5. [6] Let (X, ≼) be a partially ordered set and suppose that there exists
a metric d on X such that (X, d) is a complete metric space. Let f : X → X be a
non-decreasing mapping such that there exists x0 ∈ X with x0 ≼ f (x0 ). Suppose that
there exist β ∈ S and ψ ∈ Ψ, ψ sub-additive such that
ψ(d(f (x), f (y))) ≤ β(ψ(d(x, y)))ψ(d(x, y))
”(3)”
for all x, y ∈ X with x ≽ y.
Assume that either
(i) f is continuous (or)
(ii) X is such that if an increasing sequence {xn } converges to x then xn ≼ x for each
n≥1
holds. Then f has a fixed point.
Further, if for each x, y ∈ X, there exists z ∈ X which is comparable to x and y, then f
has a unique fixed point in X.
Here we note that ψ ◦ d = ρ (say) is a metric if ψ ∈ Ψ is sub-additive. It is clear
that (X, ρ) is complete if and only if (X, d) is complete. In this case by choosing ρ = ψod
in (3) of Theorem 1.5 then the inequality (3) reduces to (2) of Theorem 1.4 with ρ in
place of d and the conclusion of Theorem 1.4 holds. Hence there is no importance of ψ in
Theorem 1.5. So in order to show the importance of ψ ∈ Ψ, we relax the sub-additivity
property of ψ of Theorem 1.5 and prove the existence of fixed points by using a more
general Geraghty contraction than that of (3).
Definition 1.6. [10] A selfmap f : X → X, where (X, d) is a metric space is called a
Kannan type mapping if there exists 0 < λ < 1 such that
”(4)”
d(f (x), f (y)) ≤ λ2 (d(x, f (x)) + d(y, f (y)))
for all x, y ∈ X.
In 2013, Choudhury and Kundu [4] extended Theorem 1.2 to Kannan type mappings
in partially ordered metric spaces.
Theorem 1.7. [4] Let (X, ≼) be a partially ordered set and suppose that there exists
a metric d on X such that (X, d) is a complete metric space. Let f : X → X be a
non-decreasing mapping such that there exists x0 ∈ X with x0 ≼ f (x0 ). Suppose that
there exists β ∈ S such that
d(f (x), f (y)) ≤ β( 12 (d(x, f (x)) + d(y, f (y))))( 21 (d(x, f (x)) + d(y, f (y))))
”(5)”
for all x, y ∈ X, x and y are comparable.
Also suppose that either
(i) f is continuous (or)
(ii) X has the property, if an increasing sequence {xn } converges to x then xn ≼ x for
each n ≥ 0.
Then f has a fixed point in X.
12
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
Theorem 1.8. [4] In addition to the hypotheses of Theorem 1.7, assume that for any
y, z ∈ X, there exists w ∈ X which is comparable to y and z and is such that w ≼ f (w),
then the fixed point in Theorem 1.7 is unique.
Remark 1.9. A mapping f : X → X that satisfies the inequality (5) is said to be
Geraghty contraction of Kannan type.
The study of the existence of fixed points of weak Geraghty contractions in partially
ordered metric spaces by using an altering distance function is of present interest. In
this direction, we introduce ψ-weak Geraghty contractions and prove the existence of
fixed points in partially ordered complete metric spaces by using an altering distance
function. Examples are provided to show that our results generalize the results of
Gordji, Ramezani, Cho and Pirbavafa [6], Choudhury and Kundu [4].
2
ψ-Weak generalized Geraghty contractions
In the following, first we introduce generalized Geraghty contractions in partially
ordered metric spaces. Throughout this section, we assume that (X, ≼) is a partially
ordered set and there exists a metric d on X such that (X, d) is a metric space. Let
f : X → X be a selfmap of X.
Definition 2.1. We say that f is a ‘generalized Geraghty contraction’ if there exists
β ∈ S such that
d(f (x), f (y)) ≤ β(M (x, y))M (x, y))
”(6)”
1
1
where M (x, y) = max{d(x, y), 2 (d(x, f (x)) + d(y, f (y))), 2 (d(x, f (y)) + d(y, f (x)))}
for any comparable elements x, y ∈ X.
Example 2.1. Let X = {0, 1, 2, 3, 4} be endowed with the usual metric d. We define
partial order ≼ on X as follows:
≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (0, 4), (1, 4)}.
Clearly (X, d) is a metric space and (X, ≼) is a partially ordered set.
We define f : X → X by f (0) = 2, f (1)
= 2, f (2) = 3, f (3) = 4 and f (4) = 4.
if t > 1
 34
1 − t if 0 < t ≤ 1
We define β : [0, ∞) → [0, 1) by β(t) =

0
if t = 0.
We now verify the inequality (6).
Case (i): (x, y) = (0 , 4).
In this case, d(f (0), f (4)) = 2, M (0, 4) = 4, and
d(f (0), f (4)) = 2 ≤ 34 .4 = β(M (0, 4)).M (0, 4)
so that the inequality (6) holds.
Case (ii): (x, y) = (1, 4).
Here d(f (1), f (4)) = 2, M (1, 4) = 3, and
d(f (1), f (4)) = 2 ≤ 34 .3 = β(M (1, 4)).M (1, 4)
so that the inequality (6) holds.
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
13
In the remaining cases (6) holds trivially.
Hence f is a generalized Geraghty contraction.
The following example shows that every generalized Gerghty contraction need not
be a Geraghty contraction.
Example 2.2. Let X = {0, 1, 2, 3, 4, 5} be endowed with the usual metric d. We define
the partial ordering on X as follows:
≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (0, 4), (2, 3)}.
Clearly (X, d) is a metric space and (X, ≼) is a partially ordered set.
We define f : X → X by f (0) = 
2, f (1) = 2, f (2) = 5, f (3) = 4, f (4) = 3 and f (5) = 5
if t > 1
 12
1 − t if 0 < t ≤ 1
and β : [0, ∞) → [0, 1) by β(t) =

0
if t = 0.
We now verify the inequality (6).
Case (i): (x, y) = (0, 4).
In this case d(f (0), f (4)) = 1, M (0, 4) = 4, and
d(f (0), f (4)) = 1 ≤ 12 .4 = β(M (0, 4)).M (0, 4)
so that the inequality (6) holds.
Case (ii): (x, y) = (2, 3).
Now d(f (2), f (3)) = 1, M (2, 3) = 2, and
d(f (2), f (3)) = 1 ≤ 12 .2 = β(M (2, 3)).M (2, 3).
Therefore the inequality (6) holds.
In the remaining cases (6) holds trivially.
Hence f is a generalized Geraghty contraction.
Now, for (x, y) = (2, 3), we have
d(f (2), f (3)) = 1, d(2, 3) = 1, and
1 = d(f (2), f (3)) β(d(2, 3)).d(2, 3) = β(1).1 for any β ∈ S so that f is not a Geraghty
contraction.
Here we note that since β ∈ S has no monotonicity property, we can not conclude
that every Geraghty contraction is a generalized Geraghty contraction.
We now introduce ψ- generalized Geraghty contraction in partially ordered metric
spaces.
Definition 2.2. Let ψ ∈ Ψ. If there exists β ∈ S satisfying
ψ(d(f (x), f (y))) ≤ β(ψ(M (x, y)))ψ(M (x, y))
”(7)”
where M (x, y) = max{d(x, y), 21 (d(x, f (x)) + d(y, f (y))), 21 (d(x, f (y)) + d(y, f (x)))}
for any comparable elements x, y ∈ X, then we say that f is a ψ-generalized Geraghty
contraction.
Here we note that if ψ is the identity map in the inequality (7) then it is a
generalized Geraghty contraction.
14
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
Example 2.3. Let X = {1, 2, 3} with the usual metric d. We define partial order ≼ on
X as follows:
≼:= {(1, 1), (2, 2), (3, 3), (1, 3), (2, 3)}.
We define f : X → X byf (1) = 3, f (2) = 2 and f (3) = 2.
We define ψ : [0, ∞) → [0, ∞) by ψ(t)= 4t , and
{ −t
e
if t > 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
We now verify the inequality (7).
Case (i): (x, y) = (1, 3).
In this case, d(f (1), f (3)) = 1, ψ(d(f (1), f (3))) = 14 , M (1, 3) = 2, ψ(M (1, 3)) = 21 , and
ψ(d(f (1), f (3))) = 41 ≤ 11 . 12 = β(ψ(M (1, 3))).ψ(M (1, 3))
e2
so that the inequality (7) holds.
In the remaining cases the inequality (7) holds trivially. Hence f is a ψ generalized
Geraghty contraction.
Definition 2.3. A selfmap f on X is said to be a weak generalized Geraghty contraction
if there exist β ∈ S and L ≥ 0 such that
d(f (x), f (y)) ≤ β(M (x, y))M (x, y)) + L.N (x, y)
”(8)”
1
1
where M (x, y) = max{d(x, y), 2 (d(x, f (x)) + d(y, f (y))), 2 (d(x, f (y)) + d(y, f (x)))}
N (x, y) = min{d(x, f (x), d(x, f (y), d(y, f (x)}
for all x, y ∈ X with x ≽ y.
Here we observe that every generalized Geraghty contraction is a weak generalized
Geraghty contraction. But its converse need not be true.
Example 2.4. Let X = {0, 1, 2, 3} with the usual metric d. We define partial order ≼
on X as follows: ≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2), (1, 3)}.
We define f : X → X by f (0){= 0, f (1) = 1, f (2) = 3 and f (3) = 2, and
1
1+t if t ≥ 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
We now verify the inequality (8) for the elements (1, 2) and (1, 3) with L = 2 and in the
remaining cases the inequality (8) holds trivially.
Case (i): (x, y) = (2, 1).
In this case, d(f (2), f (1)) = 2, M (2, 1) = 32 , N (2, 1) = 1, and
d(f (2), f (1)) = 2 ≤ 35 + L.1 = β(M (2, 1)).M (2, 1) + L.N (2, 1) so that the inequality (8)
holds with L = 2.
Case (ii): (x, y) = (1, 3).
Here d(f (3), f (1)) = 1, M (3, 1) = 2, N (3, 1) = 1, and
d(f (3), f (1)) = 1 ≤ 23 + L.1 = β(M (3, 1)).M (3, 1) + L.N (3, 1) so that the inequality (8)
holds with L = 2.
Hence f is a weak generalized Geraghty contraction with L = 2, but f is not a generalized
Geraghty contraction.
Definition 2.4. Let ψ ∈ Ψ. If there exist β ∈ S and L ≥ 0 such that
ψ(d(f (x), f (y))) ≤ β(ψ(M (x, y)))ψ(M (x, y)) + L.N(x, y)
”(9)”
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
15
where M (x, y) = max{d(x, y), 21 (d(x, f (x)) + d(y, f (y))), 12 (d(x, f (y)) + d(y, f (x)))}
N (x, y) = min{d(x, f (x)), d(x, f (y)), d(y, f (x))} for all x, y ∈ X with x ≽ y
then we call f is a ψ-weak generalized Geraghty contraction.
Example 2.5. Let X = {0, 1, 2, 3, 4} with the usual metric d. We define partial order
≼ on X as follows: ≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (0, 2), (0, 3), (1, 2), (1, 3)}.
We define f : X → X by f (0) = 0, f (1) = 0, f (2) = 3, f (3) = 4 and f (4) = 4.
Define ψ : [0, ∞) → [0, ∞) by{ψ(t) = t2 , t≥ 0 and
1
1+t if t > 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
We now verify the inequality (9) for the elements (0, 2), (0, 3), (1, 2) and (1, 3) with L = 16
where as in the remaining cases the inequality (9) holds trivially.
Case(i): (x, y) = (2, 0)
In this case, ψ(d(f (2), f (0))) = 9, M (2, 0) = 52 , N (2, 0) = 1, and
25
ψ(d(f (2), f (0))) = 9 ≤ 29
+ L.1 = β(ψ(M (2, 0))).ψ(M (2, 0)) + L.N (2, 0).
Case(ii): (x, y) = (3, 0)
Here ψ(d(f (3), f (0))) = 16, M (3, 0) = 72 , N (3, 0) = 1, and
ψ(d(f (3), f (0))) = 16 ≤ 49
53 + L.1 = β(ψ(M (3, 0))).ψ(M (3, 0)) + L.N (3, 0).
Case(iii): (x, y) = (2, 1).
Now ψ(d(f (2), f (1))) = 9, M (2, 1) = 2, N (2, 1) = 1, and
ψ(d(f (2), f (1))) = 9 ≤ 45 + L.1 = β(ψ(M (2, 1))).ψ(M (2, 1)) + L.N (2, 1).
Case(iv): (x, y) = (1, 3).
Here ψ(d(f (3), f (1))) = 16, M (3, 1) = 3, N (3, 1) = 1, and
9
ψ(d(f (3), f (1))) = 16 ≤ 10
+ L.1 = β(ψ(M (3, 1))).ψ(M (3, 1)) + L.N (3, 1).
Therefore f is a ψ-weak generalized Geraghty contraction with L = 16.
The following lemma is useful in proving our main results.
Lemma 2.5. [2] Let (X, d) be metric space. Let {xn } be a sequence in X such that
d(xn+1 , xn ) → 0 as n → ∞. If {xn } is not a Cauchy sequence then there exist an ϵ > 0
and sequences of positive integers {m(k)} and {n(k)} with m(k) > n(k) > k and
(i) lim d(xm(k)−1 , xn(k)+1 ) = ϵ, (ii) lim d(xm(k) , xn(k) ) = ϵ,
k→∞
k→∞
(iii) lim d(xm(k)−1 , xn(k) ) = ϵ, and (iv) lim d(xm(k) , xn(k)+1 ) = ϵ.
k→∞
3
k→∞
Main results
Theorem 3.1. Let (X, ≼) be a partially ordered set and suppose that d is a metric on
X such that (X, d) is a complete metric space. Let f : X → X be a non-decreasing
mapping such that there exists x0 ∈ X with x0 ≼ f (x0 ). Assume that f is ψ-weak
generalized Geraghty contraction.
Furthermore, assume that either
(i) f is continuous (or)
16
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
(ii) X is such that if {xn } ⊂ X is a non-decreasing sequence with xn → x, then xn ≼ x
for all n ≥ 1.
Further, if for any s > 0, lim sup β(t) = β(s) then f has a fixed point in X.
t→s
Proof. We define {xn } in X by xn = f n (x0 ) = f (xn−1 ) for each n ≥ 1.
Since x0 ≼ f (x0 ) and f is a non-decreasing function, by mathematical induction
it follows that
x0 ≼ f (x0 ) ≼ f 2 (x0 ) ≼ f 3 (x0 ) ≼ ... ≼ f n (x0 ) ≼ f n+1 (x0 ) ≼ ...
so that xn ≼ xn+1 for each n ≥ 0.
If xn = xn+1 for some n then xn is a fixed point of f .
Without loss of generality, we assume that xn ̸= xn+1 for each n.
We have
ψ(d(xn+2 , xn+1 )) = ψ(d(f n+2 (x0 ), f n+1 (x0 )))
= ψd(f (xn+1 , f (xn )))
≤ β(ψ(M (xn+1 , xn )))ψ(M (xn+1 , xn )) + LN (xn+1 , xn ).
”(10)”
Now
M (xn+1 , xn ) = max{d(xn+1 , xn ), 12 (d(xn+1 , f (xn+1 )) + d(xn , f (xn ))),
1
2 (d(xn+1 , f (xn )) + d(xn , f (xn+1 )))}
= max{d(xn+1 , xn ), 12 (d(xn+1 , xn+2 ) + d(xn , xn+1 )),
1
2 (d(xn+1 , xn+1 ) + d(xn , xn+2 ))}
= max{d(xn+1 , xn ), 12 (d(xn+1 , xn+2 ) + d(xn , xn+1 )), 12 d(xn , xn+2 )}
≤ max{d(xn+1 , xn ), 21 (d(xn+1 , xn+2 ) + d(xn , xn+1 )),
1
2 (d(xn , xn+1 ) + d(xn+1 , xn+2 ))}
≤ max{d(xn+1 , xn ), 12 (d(xn+1 , xn+2 ) + d(xn , xn+1 ))}
≤ max {d(xn+1 , xn ), d(xn+1 , xn+2 ) }.
Also,
N (xn+1 , xn ) = min{d(xn+1 , f (xn+1 ), d(xn+1 , f (xn )), d(xn , f (xn+1 )}
= min{d(xn+1 , xn+2 ), d(xn+1 , xn+1 ), d(xn , xn+2 )}
= min{d(xn+1 , xn+2 ), 0, d(xn , xn+2 )} = 0.
If max{d(xn+1 , xn ), d(xn+1 , xn+2 )} = d(xn+1 , xn+2 )
then from (10), we have
ψ(d(xn+2 , xn+1 )) ≤ β(ψ(M (xn+1 , xn )))ψ(M (xn+1 , xn ))
≤ β(ψ(M (xn+1 , xn )))ψ(d(xn+2 , xn+1 ))
< ψ(d(xn+2 , xn+1 )),
a contradiction.
So we have max{d(xn+1 , xn ), d(xn+1 , xn+2 )} = d(xn+1 , xn ), and hence
ψ(d(xn+2 , xn+1 )) ≤ β(ψ(M (xn+1 , xn )))ψ(d(xn+1 , xn )) + L.0
< ψ(d(xn+1 , xn )) for all n.
Thus it follows that {ψ(d(xn+1 , xn ))} is a decreasing sequence of non-negative reals and
so lim ψ(d(xn+1 , xn )) exists and it is r (say). i.e., lim ψ(d(xn+1 , xn )) = r ≥ 0.
n→∞
n→∞
We now show that r = 0.
If r > 0 then from (10) we have
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
17
ψ(d(xn+2 , xn+1 )) = ψ(d(f xn+1 , f xn ))
≤ β(ψ(M (xn+1 , xn )))ψ(M (xn+1 , xn )) + LN (xn+1 , xn )
= β(ψ(M (xn+1 , xn )))ψ(M (xn+1 , xn ))
≤ β(ψ(M (xn+1 , xn )))ψ(d(xn+1 , xn )), and hence
ψ(d(xn+2 ,xn+1 ))
ψ(d(xn+1 , xn )) ≤ β(ψ(M (xn+1 , xn ))) < 1 for each n ≥ 1.
Now on letting n → ∞, we get
n+2 ,xn+1 ))
1 = lim ψ(d(x
ψ(d(xn+1 ,xn )) ≤ lim β(ψ(M (xn+1 , xn ))) ≤ 1
n→∞
n→∞
so that β(ψ(M (xn+1 , xn ))) → 1 as n → ∞.
This implies that lim (ψ(M (xn+1 , xn ))) = 0.
n→∞
Since ψ(d(xn+1 , xn )) ≤ ψ(M (xn+1 , xn )) for all n, we have
lim (ψ(d(xn+1 , xn ))) ≤ lim (ψ(M (xn+1 , xn ))) = 0.
n→∞
n→∞
Hence lim ψ(d(xn+1 , xn )) = 0. i.e., r = 0.
n→∞
Suppose that {xn } is not a Cauchy sequence. Then by Lemma 2.5, there exist an
ϵ > 0 and sequences of positive integers {m(k)} and {n(k)} with m(k) > n(k) > k and
(i), (ii), (iii) and (iv) of Lemma 2.5 hold.
By taking x = xn(k) , y = xm(k)−1 in (9), it follows that
ψ(d(xn(k)+1 , xm(k) ) = ψ(d(f (xn(k) ), f (xm(k)−1 )))
≤ β(ψ(M (xn(k) , xm(k)−1 )))ψ(M (xn(k) , xm(k)−1 )) + LN (xn(k) , xm(k)−1 )
where
M (xn(k) , xm(k)−1 ) = max{d(xn(k) , xm(k)−1 ), 12 (d(xn(k) , f (xn(k) ))+d(xm(k)−1 , f (xm(k)−1 ))),
1
2 (d(xn(k) , f (xm(k)−1 )) + d(xm(k)−1 , f (xn(k) )))}
= max{d(xn(k) , xm(k)−1 ), 21 (d(xn(k) , xn(k)+1 ) + d(xm(k)−1 , xm(k) )),
1
2 (d(xn(k) , xm(k) ) + d(xm(k)−1 , xn(k)+1 ))}.
On letting k → ∞ and from the Lemma 2.5, we get
lim M (xn(k) , xm(k)−1 ) = max{ϵ, 0, ϵ} = ϵ.
k→∞
Also, we have
N (xn(k) , xm(k)−1 ) = min{d(xn(k) , f (xn(k) )), d(xn(k) , f (xm(k)−1 )), d(xm(k)−1 , f (xn(k) ))}
= min{d(xn(k) , xn(k)+1 ), d(xn(k) , xm(k) ), d(xm(k)−1 , xn(k)+1 )}.
On letting k → ∞ and from the Lemma 2.5, we get
lim N (xn(k) , xm(k)−1 ) = min{0, ϵ, ϵ} = 0.
k→∞
Now, we have
ψ(d(xn(k)+1 , xm(k) )) ≤ β(ψ(M (xn(k) , xm(k)−1 )))ψ(M (xn(k) , xm(k)−1 ))+LN (xn(k) , xm(k)−1 )
≤ β(ψ(M (xn(k) , xm(k)−1 )))ψ(M (xn(k) , xm(k)−1 ))
≤ β(ψ(M (xn(k) , xm(k)−1 )))ψ(d(xn(k) , xm(k)−1 )), and hence
ψ(d(xn(k)+1 ,xm(k) ))
ψ(d(xn(k) ,xm(k)−1 ))
≤ β(ψ(M (xn(k) , xm(k)−1 ))) < 1.
On letting k → ∞ and from the Lemma 2.5, we get
1 = ψ(ϵ)
ψ(ϵ) ≤ lim β(ψ(M (xn(k) , xm(k)−1 ))) ≤ 1
k→∞
so that β(ψ(M (xn(k) , xm(k)−1 ))) → 1 as k → ∞.
Since β ∈ S, ψ(M (xn(k) , xm(k)−1 )) → 0 as k → ∞. i.e., ψ(ϵ) = 0,
since ψ is continuous. Hence it follows that ϵ = 0,
18
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
a contradiction.
Therefore {xn } is a Cauchy sequence in X, and since X is complete, there exists u ∈ X
such that lim xn = u.
n→∞
Now, we show that u is a fixed point of f .
First we assume that (i) holds. i.e., f is continuous.
In this case, we have u = lim f n (x0 ) = lim f n+1 (x0 ) = f ( lim f n (x0 )) = f (u).
n→∞
n→∞
n→∞
Therefore u is a fixed point of f in X.
We now assume that (ii) holds. Since {xn } ⊂ X is an increasing sequence in X and
xn → u as n → ∞, by (ii) we have xn ≼ u for each n ≥ 1.
We consider
ψ(d(f (u), f (xn ))) ≤ β(ψ(M (u, xn )))ψ(M (u, xn )) + L.N (u, xn )
”(11)”
M (u, xn ) = max{d(u, xn ), 12 (d(u, f (u)) + d(xn , f (xn ))), 21 (d(u, f (xn )) + d(xn , f (u)))}
= max{(d(u, xn ), 21 (d(u, f (u)) + d(xn , xn+1 )), 21 (d(u, (xn+1 ) + d(xn , f (u)))}.
On letting n → ∞, we get
lim M (u, xn ) = max{d(u, u), 21 (d(u, f (u)) + d(u, u)), 21 (d(u, u) + d(u, f (u)))}
n→∞
= 12 d(u, f (u)).
Also,
N (u, xn ) = min{d(u, f (u)), d(u, f (xn )), d(xn , f (u))}
= min{d(u, f (u)), d(u, xn+1 ), d(xn , f (u))}.
On letting n → ∞, we have
lim N (u, xn ) = min{d(u, f (u)), d(u, u), d(u, f (u))} = min{d(u, f (u)), 0)} = 0.
n→∞
From (11), we have
lim sup ψ(d(f (u), f (xn )) ≤ lim sup β(ψ(M (u, xn ))) lim sup ψ(M (u, xn ))+lim sup L.N (u, xn ).
n→∞
n→∞
n→∞
Hence it follows that ψ(d(f (u), u)) ≤ β(ψ( 21 d(u, f (u))))ψ( 21 d(u, f (u)))
which implies that d(f (u), u) ≤ 21 (d(u, f (u)))
so that d(f (u), u) = 0. Hence u is a fixed point of f in X.
n→∞
Theorem 3.2. In addition to the hypotheses of Theorem 3.1 we assume the following:
“for every u, v ∈ X, there exists x ∈ X which is comparable to u and v”.
”(12)”
Then f has a unique fixed point in X.
Proof. Suppose that there exist u, v ∈ X which are fixed points of f .
The following two cases are possible.
Case (i): Suppose that u is comparable to v.
Then f n (u) = u is comparable to f n (v) = v for n = 0, 1, 2, 3, ... and
ψ(d(u, v)) = ψ(d(f n (u), f n (v)))
= ψ(d(f (f n−1 (u)), f (f n−1 (v))))
≤ β(ψ(M (f n−1 (u), f n−1 (v))))ψ(M (f n−1 (u), f n−1 (v)))+L.N (f n−1 (u), f n−1 (v))
where
M (f n−1 (u), f n−1 (v)) = max{d(f n−1 (u), f n−1 (v)), 21 (d(f n−1 (u), f n (u))+d(f n−1 (v), f n (v))),
1
n−1 (u), f n (v)) + d(f n−1 (v), f n (u)))}
2 (d(f
= max{d(u, v), 21 (d(u, u) + d(f (v), v)), 21 (d(u, v) + d(f (v), u))}
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
19
= max{d(u, v), 21 (d(u, u) + d(v, v)), 21 (d(u, v) + d(v, u))}
= d(u, v) and
n−1
n−1
N (f
(u), f
(v)) = min{d(f n−1 (u), f n (u)), d(f n−1 (u), f n (v)), d(f n−1 (v), f n (u))}
= min{d(u, u), d(u, v), d(v, u)} = 0.
Therefore ψ(d(u, v)) ≤ β(ψ(d(u, v)))ψ(d(u, v)) + L.0.
Since β ∈ S, it follows that ψ(d(u, v)) = 0, so that u = v.
Case (ii): Suppose that u is not comparable to v.
In this case there exists x ∈ X which is comparable to both u and v. The monotonicity
of f implies that f n (x) is comparable to f n (u) = u and f n (v) = v for n ≥ 0.
Now
ψ(d(u, f n (x))) = ψ(d(f n (u), f n (x)))
= ψ(d(f (f n−1 (u)), f (f n−1 (x))))
≤ β(ψ(M (f n−1 (u), f n−1 (x))))ψ(M (f n−1 (u), f n−1 (x))) +
L.N (f n−1 (u), f n−1 (x))
”(13)”
where
M (f n−1 (u), f n−1 (x)) = max{d(f n−1 (u), f n−1 (x)), 12 (d(f n−1 (u), f n (u))+d(f n−1 (x), f n (x))),
1
n−1 (u), f n (x)) + d(f n−1 (x), f n (u)))}.
2 (d(f
= max{d(u, f n−1 (x)), 21 (d(u, u) + d(f n−1 (x), f n (x))),
1
n
n−1 (x), u))}.
2 (d(u, f (x)) + d(f
1
n−1
n
n−1
(x), f (x))),
= max{d(u, f
(x)), 2 (d(f
1
n
n−1 (x), u))}.
2 (d(u, f (x)) + d(f
1
n−1
n−1
(x), u) + d(u, f n (x)))}
≤ max{d(u, f
(x)), 2 (d(f
≤ max{d(u, f n−1 (x)), d(u, f n (x))}, and
n−1
n−1
N (f
(u), f
(x)) = min{d(f n−1 (u), f n (u)), d(f n−1 (u), f n (x)), d(f n−1 (x), f n (u))}
= min{d(u, u), d(u, f n (x)), d(f n−1 (x), u))}
= min{0, d(u, f n (x)), d(f n−1 (x), u))} = 0.
n−1
If max{d(u, f
(x)), d(u, f n (x))} = d(u, f n (x)), then from (13), we have
ψ(d(u, f n (x)) ≤ β(ψ(M (f n−1 (u), f n−1 (x))))ψ(M (f n−1 (u), f n−1 (x))).
Since β ∈ S it follows that ψ(d(u, f n (x))) < ψ(d(u, f n (x))),
a contradiction.
Now from (13), we have ψ(d(u, f n (x))) ≤ β(ψ(M (f n−1 (u), f n−1 (x))))ψ(d(u, f n−1 (x))).
Since β ∈ S, it follows that
ψ(d(u, f n (x))) < ψ(d(u, f n−1 (x))) for each n = 1, 2, ... .
Therefore {ψ(d(u, f n (x)))} is a non-increasing sequence of non-negative real numbers.
Thus, there exists r ≥ 0 such that lim ψ(d(u, f n (x))) = r.
n→∞
If r > 0 , then from (13) we have
ψ(d(u, f n (x))) ≤ β(ψ(M (f n−1 (u), f n−1 (x))))ψ(d(u , f n−1 (x))), and hence
ψ(d(u,f n (x)))
≤ β(ψ(M (f n−1 (u), f n−1 (x)))) < 1.
ψ(d(u,f n−1 (x)))
On letting n → ∞,
n−1 (u), f n−1 (x)))) ≤ 1
we have 1 = ψ(r)
ψ(r) ≤ lim β(ψ(M (f
n→∞
which implies that lim β(ψ(M (f n−1 (u), f n−1 (x)))) = 1.
n→∞
20
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
Since β ∈ S it follows that lim ψ(M (f n−1 (u), f n−1 (x))) = 0.
n→∞
Since ψ(d(u, f n−1 (x))) ≤ (ψ(M (f n−1 (u), f n−1 (x)))),
it follows that r = lim ψ(d(u, f n−1 (x)) = 0.
n→∞
Since ψ is continuous, we have ψ( lim d(u, f n−1 (x))) = 0.
This implies that lim f n (x) = u.
n→∞
n→∞
On the similar lines as above, we can show that lim f n (x) = v.
n→∞
Now, by the uniqueness of limits, we get u = v.
4
Corollaries and examples
The following is a corollary to Theorem 3.2
Corollary 4.1. Let (X, ≼) be a partially ordered set and suppose that d is a metric on
X such that (X, d) is a complete metric space. Let f : X → X be a non-decreasing
mapping such that there exists x0 ∈ X with x0 ≼ f (x0 ). Suppose that there exist
ψ ∈ Ψ, β ∈ S and there is a positive integer m such that
ψ(d(f m (x), f m (y))) ≤ β(ψ(M (x, y))ψ(M (x, y)) + L.N(x,y)
”(14)”
where M (x, y) = max{d(x, y), 21 (d(x, f m (x))+d(y, f m (y))), 21 (d(x, f m (y))+d(y, f m (x)))},
N (x, y) = min{d(x, f m (x)), d(x, f m (y)), d(y, f m (x))} ∀ x,y ∈ X with x ≽ y.
Assume that either
(i) f is continuous (or)
(ii) X is such that if {xn } ⊂ X is a non-decreasing sequence with xn → x, then xn ≼ x
for all n ≥ 1.
Further, assume that for every u, v ∈ X , there exists x ∈ X which is comparable to u
and v.
”(15)”
Then f has a unique fixed point in X.
Proof. By replacing f by f m in Theorem 3.2, it follows that f m has a unique fixed point
in X, say u in X. Now f (u) =f (f m (u)) = f m+1 (u) = f m (f (u)). Hence f (u) is also a
fixed point of f m .
Therefore, by the uniqueness of fixed point of f m , it follows that f (u) = u.
Now we present an example in support of Theorem 3.1.
Example 4.1. Let X = {0, 1, 2, 3, 4} with the usual metric.
We define partial order ≼ on X as follows:
≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (3, 1), (4, 1), (4, 2), (4, 3)}.
Clearly (X, d) is a metric space and (X, ≼) is a partially ordered set.
We define f : X → X by f (0) = 0, f (1) = 2, f (2) = 3, f (3) = 4 and f (4) = 4.
Then f is continuous and non-decreasing.
Moreover, we choose x0 = 0 ∈ X then x0 ≼ f (x0 ).
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
21
{
2 2
if 0 ≤ t < 1
3t
We define ψ : [0, ∞) → [0, ∞) by ψ(t) =
2
t
if
t ≥ 1, and
3
{ 1
1+t if t > 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
Then ψ ∈ Ψ and β ∈ S.
We now verify the inequality (9) for the elements (3, 1), (4, 1) and (4, 2) with L = 1, and
in the remaining cases the inequality (9) holds trivially.
Case (i) : (x, y) = (1, 3).
In this case ψ(d(f (1), f (3))) = 43 , M (1, 3) = 2, N (1, 3) = 1, and
ψ(d(f (1), f (3))) = 34 ≤ 47 + L.1 = β(ψ(M (1, 3))).ψ(M (1, 3)) + L.N (1, 3).
Case (ii) : (x, y) = (1, 4).
In this case ψ(d(f (1), f (4))) = 43 , M (1, 4) = 3, N (1, 4) = 1, and
ψ(d(f (1), f (4))) = 34 ≤ 23 + L.1 = β(ψ(M (1, 4))).ψ(M (1, 4)) + L.N (1, 4).
Case (iii) : (x, y) = (2, 4).
Now ψ(d(f (2), f (4))) = 23 , M (2, 4) = 2, N (2, 4) = 1, and
ψ(d(f (2), f (4))) = 23 ≤ 47 + L.1 = β(ψ(M (2, 4))).ψ(M (2, 4)) + L.N (2, 4).
Hence f is a ψ-weak generalized Geraghty contraction, and f satisfies all the conditions
of Theorem 3.1 and f has two fixed points 0 and 4.
Now we present an example in support of Theorem 3.2.
Example 4.2. Let X = {0, 1, 2} with the usual metric.
We define partial order ≼ on X as follows: ≼:= {(0, 0), (1, 1), (2, 2), (2, 0), (2, 1)}.
Clearly (X, d) is a metric space and (X, ≼) is a partially ordered set.
We define f : X → X by f (0) = 1, f (1) = 2 and f (2) = 2.
Then f is continuous and non-decreasing. Moreover, we choose x0 = 2 ∈ X, x0 ≼ f (x0 ).
We define ψ : [0, ∞) → [0, ∞){by ψ(t) = t2 , t≥ 0 and
1
1+t if t > 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
Then ψ ∈ Ψ and β ∈ S.
We now verify the inequality (9) for the element (2, 0), with L = 1, and in the remaining
cases the inequality (9) holds trivially.
Case(i) : (x, y) = (0, 2)
Here ψ(d(f (0), f (2))) = 1, M (0, 2) = 2, N (0, 2) = 1
ψ(d(f (0), f (2))) = 1 ≤ 45 + L.1 = β(ψ(M (0, 2))).ψ(M (0, 2)) + L.N (0, 2).
So f is a ψ-weak generalized Geraghty contraction and f satisfies (12) also. Therefore
f satisfies all the conditions of Theorem 3.2 and f has only one fixed point 2.
We now present an example in support of Corollary 4.1
Example 4.3. Let X = {0, 1, 2} with the usual metric. We define partial order ≼ on
X as follows: ≼:= {(0, 0), (1, 1), (2, 2), (1, 0), (2, 0), (2, 1)}.
We now define f : X → X by f (0) = 0, f (1) = 0, and f (2) = 1. Then f is nondecreasing.
22
Fixed points of ψ-weak Geraghty contractions in partially ordered metric spaces
{
We define ψ : [0, ∞) → [0, ∞) by ψ(t) =
t2 ,
t ≥ 0, and β(t) =
1
1+t
if t > 0
if t = 0.
0
Then ψ ∈ Ψ and β ∈ S.
We now verify the inequality (14) for the element (2, 1).
Case (i) : (x, y) = (1, 2)
In this case ψ(d(f 2 (1), f 2 (2))) = 0, hence the inequality (14) holds trivially with m = 2.
At all remaining points also the inequality (14) holds trivially with m = 2.
Further, we note that for every u and v in X there is x = 2 ∈ X which is comparable to
both u and v.
Therefore f satisfies all the conditions of Corollary 4.1 and f has only one fixed point
0.
But when m = 1, the inequality (14) fails to hold when (x, y) = (1, 2). For,
ψ(d(f (1), f (2))) = ψ(1), M (1, 2) = 1, N (1, 2) = 0, and
ψ(d(f (1), f (2))) = ψ(1) β(ψ(1)).ψ(1) + L.0 for any ψ ∈ Ψ.
Thus this is a supporting example that satisfies all the conditions of Corollary 4.1
and fails to hold for m = 1. Hence Corollary 4.1 generalizes Theorem 3.2.
Example 4.4. Let X = {0, 1, 2, 3} with the usual metric.
We define partial order ≼ on X as follows: ≼:= {(0, 0), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}.
We define f : X → X by f (0) = 0, f (1) = 1, f (2) = 1 and f (3) = 2.
Then f is non-decreasing.
We define ψ : [0, ∞) → [0, ∞){by ψ(t) = t2 , t≥ 0 and
1
1+t if t > 0
β : [0, ∞) → [0, 1) by β(t) =
0
if t = 0.
Then ψ ∈ Ψ and β ∈ S.
We now verify the inequality (14) for the elements (2, 1) and (3, 1).
When 2 ≼ 1, we have ψ(d(f 2 (1), f 2 (2))) = ψ(d(1, 1)) = 0 ; and when 3 ≼ 1 we have
ψ(d(f 2 (1), f 2 (3))) = ψ(d(1, 1)) = 0, so that the inequality (14) holds trivially with
m = 2.
Here we note that when u = 0 and v = 2, there is no x ∈ X which is comparable to
both u and v. Hence (15) is not satisfied.
Therefore f satisfies all the conditions of Corollary 4.1 except condition (15) and f
has more than one fixed point namely, 0 and 1.
But when m = 1, the inequality (14) fails to hold when (x, y) = (3, 1).
d(f (1), f (3)) = 1, M (1, 3) = 1, N (1, 3) = 0, and
ψ(d(f (1), f (3))) = ψ(1) β(ψ(1)).ψ(1) + L.0 = β(ψ(M (1, 3))).ψ(M (1, 3)) + L.N (1, 3)
for any β ∈ S, ψ ∈ Ψ and L ≥ 0.
Remark 4.2. Example 4.3 and Example 4.4 illustrate the importance of m in the
inequality (14) of the Corollary 4.1.
An open problem: Is the class of all ψ-weak generalized Geraghty contraction maps
is larger than the class of weak generalized Geraghty contraction maps?
G.V.R. Babu, K.K.M. Sarma and P.H. Krishna
23
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