Filomat 28:4 (2014), 715–722
DOI 10.2298/FIL1404715C
Published by Faculty of Sciences and Mathematics,
University of Ni
ˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Fixed Point Results for F-Contractive Mappings of Hardy-Rogers-Type
Monica Cosentino, Pasquale Vetro
a
a
Universit`a degli Studi di Palermo, Dipartimento di Matematica e Informatica, Via Archirafi, 34, 90123 Palermo, Italy
Abstract. Recently, Wardowski introduced a new concept of contraction and proved a fixed point theorem
which generalizes Banach contraction principle. Following this direction of research, in this paper, we
will present some fixed point results of Hardy-Rogers-type for self-mappings on complete metric spaces or
complete ordered metric spaces. Moreover, an example is given to illustrate the usability of the obtained
results.
1. Introduction
It is well known that the contraction mapping principle, formulated and proved in the Ph.D. dissertation
of Banach in 1920, which was published in 1922 [4], is one of the most important theorems in classical
functional analysis. Indeed it is widely considered as the source of metric fixed point theory. Also its
significance lies in its vast applicability in a number of branches of mathematics. Starting from these
considerations, the study of fixed and common fixed points of mappings satisfying a certain metrical
contractive condition attracted many researchers, see for example [1–3, 5, 6, 8–11, 16, 22–27]. The reader
can also see [18, 20, 21], for existence results of fixed points for contractive non-self-mappings.
Recently, Wardowski [28] introduced a new concept of contraction and proved a fixed point theorem
which generalizes Banach contraction principle. Following this direction of research, in this paper, we
will present some fixed point results of Hardy-Rogers-type for self-mappings on complete metric spaces or
complete ordered metric spaces. Moreover, an example is given to illustrate the usability of the obtained
results.
2. Preliminaries
The aim of this section is to present some notions and results used in the paper. Throughout the article
we denote by R the set of all real numbers, by R
+
the set of all positive real numbers and by N the set of all
positive integers.
Definition 2.1. Let F : R
+
→ R be a mapping satisfying:
(F1) F is strictly increasing;
2010 Mathematics Subject Classification. Primary 47H10; Secondary 54H25
Keywords. Metric spaces, ordered metric spaces, fixed points, F-contractions of Hardy-Rogers-type.
Received: 27 September 2013; Accepted: 02 November 2013
Communicated by Vladimir Rako
ˇ
cevi
´
c
Research supported by Universit
`
a degli Studi di Palermo, Local Project R.S. ex 60%.
Email addresses: mcosentino1@alice.it (Monica Cosentino), vetro@math.unipa.it (Pasquale Vetro)
M. Cosentino, P. Vetro / Filomat 28:4 (2014), 715–722 716
(F2) for each sequence {α
n
} ⊂ R
+
of positive numbers lim
n→+∞
α
n
= 0 if and only if lim
n→+∞
F(α
n
) = −∞;
(F3) there exists k ∈ (0, 1) such that lim
α→0
+
α
k
F(α) = 0.
We denote with F the family of all functions F that satisfy the conditions (F1)-(F3).
Definition 2.2 ([28]). Let (X, d) be a metric space. A self-mapping T on X is called an F-contraction if there exist
F ∈ F and τ ∈ R
+
such that
τ + F(d(Tx, Ty)) ≤ F(d(x, y)), (1)
for all x, y ∈ X with d(Tx, Ty) > 0.
Definition 2.3. Let (X, d) be a metric space. A self-mapping T on X is called an F-contraction of Hardy-Rogers-type
if there exist F ∈ F and τ ∈ R
+
such that
τ + F(d(Tx, Ty)) ≤ F(αd(x, y) + βd(x, Tx) + γd(y, Ty) + δd(x, Ty) + Ld(y, Tx)), (2)
for all x, y ∈ X with d(Tx, Ty) > 0, where α + β + γ + 2δ = 1, γ , 1 and L ≥ 0.
By choosing opportunely the mapping F, we obtain certain classes of contractions known in the literature,
as shown with the following examples.
Example 2.4 ([28]). Let F : R
+
→ R be given by F(x) = ln x. It is clear that F satisfies (F1)-(F2) and (F3) for any
k ∈ (0, 1). Each mapping T : X → X satisfying (1) is an F-contraction such that
d(Tx, Ty) ≤ e
−τ
d(x, y), for all x, y ∈ X, Tx , Ty.
It is clear that for x, y ∈ X such that Tx = Ty the previous inequality also holds and hence T is a contraction.
Example 2.5 ([28]). Let F : R
+
→ R be given by F(x) = ln x + x. It is clear that F satisfies (F1)-(F3). Each mapping
T : X → X satisfying (1) is an F-contraction such that
d(Tx, Ty)
d(x, y)
e
d(Tx,Ty)−d(x,y)
≤ e
−τ
, for all x, y ∈ X, Tx , Ty.
Remark 2.6. From (F1) and (1), we deduce that every F-contraction T is a contractive mapping, that is,
d(Tx, Ty) < d(x, y), for all x, y ∈ X, Tx , Ty.
From (F1) and (2), we deduce that every F-contraction of Hardy-Rogers-type T satisfies the following condition:
d(Tx, Ty)) < αd(x, y) + βd(x, Tx) + γd(y, Ty) + δd(x, Ty) + Ld(y, Tx), (3)
for all x, y ∈ X, Tx , Ty, where α + β + γ + 2δ = 1, γ , 1 and L ≥ 0.
Let X be a non-empty set. If (X, d) is a metric space and (X, ) is partially ordered, then (X, d, ) is called
an ordered metric space. Then x, y ∈ X are called comparable if x y or y x holds. Let (X, ) be a partially
ordered set. A self-mapping T on X is called non-decreasing if Tx Ty whenever x y for all x ∈ X. An
ordered metric space (X, d, ) is regular if
(r) for every non-decreasing sequence {x
n
} in X convergent to some x ∈ X, we have x
n
x for all n ∈ N ∪ {0}.
M. Cosentino, P. Vetro / Filomat 28:4 (2014), 715–722 717
3. Fixed points for F-contractions of Hardy-Rogers-type
In this section, we give some fixed point results for F-contractions of Hardy-Rogers-type in a complete
metric space.
Theorem 3.1. Let (X, d) be a complete metric space and let T be a self-mapping on X. Assume that there exist F ∈ F
and τ ∈ R
+
such that T is an F-contraction of Hardy-Rogers-type, that is,
τ + F(d(Tx, Ty)) ≤ F(αd(x, y) + βd(x, Tx) + γd(y, Ty) + δd(x, Ty) + Ld(y, Tx)), (4)
for all x, y ∈ X, Tx , Ty, where α + β + γ + 2δ = 1, γ , 1 and L ≥ 0. Then T has a fixed point. Moreover, if
α + δ + L ≤ 1, then the fixed point of T is unique.
Proof. Let x
0
∈ X be an arbitrary point, and let {x
n
} be the Picard sequence with initial point x
0
, that is,
x
n
= T
n
x
0
= Tx
n−1
. If x
n
= x
n−1
for some n ∈ N, then x
n
is a fixed point of T. Now, let d
n
= d(x
n
, x
n+1
) for all
n ∈ N ∪ {0}. If x
n
, x
n+1
, that is, Tx
n−1
, Tx
n
for all n ∈ N, using the contractive condition (4) with x = x
n−1
and y = x
n
, we get
τ + F(d
n
) = τ + F(d(x
n
, x
n+1
)) = τ + F(d(Tx
n−1
, Tx
n
))
≤ F(αd(x
n−1
, x
n
) + βd(x
n−1
, Tx
n−1
) + γd(x
n
, Tx
n
) + δd(x
n−1
, Tx
n
) + Ld(x
n
, Tx
n−1
))
= F(αd(x
n−1
, x
n
) + βd(x
n−1
, x
n
) + γd(x
n
, x
n+1
) + δd(x
n−1
, x
n+1
) + Ld(x
n
, x
n
))
= F(αd
n−1
+ βd
n−1
+ γd
n
+ δd(x
n−1
, x
n+1
))
≤ F((α + β)d
n−1
+ γd
n
+ δ[d
n−1
+ d
n
])
= F((α + β + δ)d
n−1
+ (γ + δ)d
n
).
Since F is strictly increasing, we deduce
d
n
< (α + β + δ)d
n−1
+ (γ + δ)d
n
and hence
(1 − γ − δ)d
n
< (α + β + δ)d
n−1
, for all n ∈ N.
From α + β + γ + 2δ = 1 and γ , 1, we deduce that 1 − γ − δ > 0 and so
d
n
<
α + β + δ
1 − γ − δ
d
n−1
= d
n−1
, for all n ∈ N.
Consequently,
τ + F(d
n
) ≤ F(d
n−1
), for all n ∈ N.
This implies
F(d
n
) ≤ F(d
n−1
) − τ ≤ · · · ≤ F(d
0
) − nτ, for all n ∈ N (5)
and so lim
n→+∞
F(d
n
) = −∞. By the properties (F2), we get that d
n
→ 0 as n → +∞.
Now, let k ∈ (0, 1) such that lim
n→+∞
d
k
n
F(d
n
) = 0. By (5), the following holds for all n ∈ N:
d
k
n
F(d
n
) − d
k
n
F(d
0
) ≤ d
k
n
(F(d
0
) − nτ) − d
k
n
F(d
0
) = −nτ d
k
n
≤ 0. (6)
Letting n → +∞ in (6), we deduce lim
n→+∞
n d
k
n
= 0 and hence lim
n→+∞
n
1/k
d
n
= 0. Now, lim
n→+∞
n
1/k
d
n
= 0
ensures that the series
P
+∞
n=1
d
n
is convergent. This implies that {x
n
} is a Cauchy sequence. As X is a complete
metric space there exists z ∈ X such that x
n
→ z. If z = Tz the proof is finished. Assume that z , Tz. If
Tx
n
= Tz for infinite values of n ∈ N ∪ {0}, then the sequence {x
n
} has a subsequence that converges to Tz
M. Cosentino, P. Vetro / Filomat 28:4 (2014), 715–722 718
and the uniqueness of the limit implies z = Tz. Then we can assume that Tx
n
, Tz for all n ∈ N ∪ {0}. Now,
by (3), we have
d(z, Tz) ≤ d(z, x
n+1
) + d(Tx
n
, Tz)
< d(z, x
n+1
) + αd(x
n
, z) + βd(x
n
, Tx
n
) + γd(z, Tz) + δd(x
n
, Tz) + Ld(z, Tx
n
)
= d(z, x
n+1
) + αd(x
n
, z) + βd(x
n
, x
n+1
) + γd(z, Tz) + δd(x
n
, Tz) + Ld(z, x
n+1
).
Letting n → +∞ in the previous inequality, we get
d(z, Tz) ≤ (γ + δ)d(z, Tz) < d(z, Tz),
which is a contradiction and hence z = Tz.
Now, we prove the uniqueness of the fixed point. Assume that w ∈ X is another fixed point of T,
different from z. This means that d(z, w) > 0. Taking x = z and y = w in (4), we have
τ + F(d(z, w)) = τ + F(d(Tz, Tw))
≤ F(αd(z, w) + βd(z, Tz) + γd(w, Tw) + δd(z, Tw) + Ld(w, Tz))
= F((α + δ + L)d(z, w)),
which is a contradiction, if α + δ + L ≤ 1, and hence z = w.
As a first corollary of Theorem 3.1, taking α = 1 and β = γ = δ = L = 0, we obtain Theorem 2.1 of
Wardowski [28]. Further, putting α = δ = L = 0 and β + γ = 1 and β , 0, we obtain the following version of
Kannan’s result [12].
Corollary 3.2. Let (X, d) be a complete metric space and let T be a self-mapping on X. Assume that there exist F ∈ F
and τ ∈ R
+
such that
τ + F(d(Tx, Ty)) ≤ F(βd(x, Tx) + γd(y, Ty)),
for all x, y ∈ X, Tx , Ty, where β + γ = 1, γ , 1. Then T has a unique fixed point in X.
A version of the Chatterjea [7] fixed point theorem is obtained from the Theorem 3.1 putting α = β = γ = 0
and δ = 1/2.
Corollary 3.3. Let (X, d) be a complete metric space and let T be a self-mapping on X. Assume that there exist F ∈ F
and τ ∈ R
+
such that
τ + F(d(Tx, Ty)) ≤ F(
1
2
d(x, Ty) + Ld(y, Tx)),
for all x, y ∈ X, Tx , Ty. Then T has a fixed point in X. If L ≤ 1/2, then the fixed point of T is unique.
Finally, if we choose δ = L = 0, we obtain a Reich [19] type theorem.
Corollary 3.4. Let (X, d) be a complete metric space and let T be a self-mapping on X. Assume that there exist F ∈ F
and τ ∈ R
+
such that
τ + F(d(Tx, Ty)) ≤ F(αd(x, y) + βd(x, Tx) + γd(y, Ty)),
for all x, y ∈ X, Tx , Ty, where α + β + γ = 1, γ , 1. Then T has a unique fixed point in X.
M. Cosentino, P. Vetro / Filomat 28:4 (2014), 715–722 719
4. Fixed points in ordered metric spaces
The existence of fixed points of self-mappings defined on certain type of ordered sets plays an important
role in the order theoretic approach. It has been initiated in 2004 by Ran and Reurings [17], and further
studied by Nieto and Rodr
´
ıguez-Lopez [13]. Then, several interesting and valuable results have appeared
in this direction [1, 14–16, 24].
Theorem 4.1. Let (X, d, ) be an ordered complete metric space and let T be a non-decreasing self-mapping on X.
Assume that there exist F ∈ F and τ ∈ R
+
such that T is an ordered F-contraction of Hardy-Rogers-type, that is,
τ + F(d(Tx, Ty)) ≤ F(αd(x, y) + βd(x, Tx) + γd(y, Ty) + δd(x, Ty) + Ld(y, Tx)), (7)
for all comparable x, y ∈ X, Tx , Ty, where α + β + γ + 2δ = 1, γ , 1 and L ≥ 0. If the following conditions are
satisfied:
(i) there exists x
0
∈ X such that x
0
Tx
0
;
(ii) X is regular;
then T has a fixed point. Moreover, if α + δ + L ≤ 1, then the set of fixed points of T is well ordered if and only if T
has a unique fixed point.
Proof. Let x
0
∈ X be such that x
0
Tx
0
, and let {x
n
} be the Picard sequence of initial point x
0
, that is,
x
n
= T
n
x
0
= Tx
n−1
. If x
n
= x
n−1
for some n ∈ N, then x
n
is a fixed point of T. Now, let d
n
= d(x
n
, x
n+1
) for all
n ∈ N ∪ {0}. Assume that x
n
, x
n−1
for all n ∈ N. As T is non-decreasing and x
0
Tx
0
, we deduce that
x
0
≺ x
1
≺ · · · ≺ x
n
≺ · · · , (8)
that is, x
n
and x
n+1
are comparable and Tx
n−1
, Tx
n
for all n ∈ N ∪ 0.
Proceeding as in the proof of Theorem 3.1, we obtain that {x
n
} is a Cauchy sequence. As X is a complete
metric space there exists z ∈ X such that x
n
→ z. If z = Tz the proof is finished. Assume that z , Tz. Since
X is regular, from (8), we deduce that x
n
and z are comparable and Tx
n
, Tz for all n ∈ N ∪ {0}.
Now, using (3), we obtain
d(z, Tz) ≤ d(z, x
n+1
) + d(Tx
n
, Tz)
< d(z, x
n+1
) + αd(x
n
, z) + βd(x
n
, x
n+1
) + γd(z, Tz) + δd(x
n
, Tz) + Ld(z, x
n+1
).
Letting n → +∞ in the previous inequality, we get
d(z, Tz) ≤ (γ + δ)d(z, Tz) < d(z, Tz),
which is a contradiction and hence z = Tz.
Now, we assume that α + δ + L ≤ 1 and that the set of fixed points of T is well ordered. We claim that
the fixed point of T is unique. Assume on the contrary that there exists another fixed point w in X such that
z , w. Using the condition (7), with x = z and y = w, we get
τ + F(d(z, w)) = τ + F(d(Tz, Tw))
≤ F(αd(z, w) + βd(z, Tz) + γd(w, Tw) + δd(z, Tw) + Ld(w, Tz))
= F((α + δ + L)d(z, w))
≤ F(d(z, w)),
which is a contradiction and hence z = w. Conversely, if T has a unique fixed point, then the set of fixed
points of T, being a singleton, is well ordered.
Theorem 4.2. Let (X, d, ) be an ordered complete metric space and let T be a non-decreasing self-mapping on X.
Assume that there exist F ∈ F and τ ∈ R
+
such that T is an ordered F-contraction of Hardy-Rogers-type. If the
following conditions are satisfied: