EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 3, 2018, 702-716
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Fixed point results for geraghty type generalized
F-contraction for weak α−admissible mappings in

metric-like spaces

Haitham Qawaqneh1,∗, Mohd Selmi Noorani1, Wasfi Shatanawi2,3

1 School of mathematical Sciences, Faculty of Science and Technology,Universiti Kebangsaan
Malaysia, 43600 UKM, Selangor Darul Ehsan, Malaysia
2 Department of Mathematics, Hashemite University, Zarqa 1315, Jordan
3 Department of Mathematics and General Courses, Prince Sultan University, Riyadh, Saudi
Arabia

Abstract. In this paper, we establish the existence of some fixed point results for generalized
(α, β, F )-Geraghty contraction in metric-like spaces. We provide an example in order to support
our results where some consequence applications of such result will be considered in this article. The
obtained results improve and extend some well-known common fixed point results in the literature.

2010 Mathematics Subject Classifications: 47H10,54H25

Key Words and Phrases: Fixed point, Metric-like space, α−admissible mapping, Weak α−admissible
mapping, F− contraction.

1. Introduction and Preliminaries

During the last decades, issues related to ”Fixed Point Theory” in order to semantics
domain with a notion of distance that has been extensively researched in different spaces.
Recently, different generalizations of metric spaces have been introduced (for example see
[12],[10],[22],[2],[28],[8],[10],[7],[6],[23],[27],[29],[32]). In 1994, Matthews [19] introduced the
notion of partial metric space as a part of the study of denotational semantics of dataflow
networks, showing that the contraction mapping principle [9] can be generalized to the
partial metric context for applications in program verifications. Later on, there have been
several recent extensive researches on (common) fixed points for different contractions on
partial metric spaces, see [[10],[1],[17],[1],[30],[24],[16],21,[3],[5], [11],[13],[25],[15],[20],[4]].

In this section, we recall some basic definitions and concepts.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3294

Email addresses: haitham.math77@gmail.com (H. Qawaqneh), msn@ukm.my (M.S. Noorani),
swasfi@hu.edu.jo (W. Shatanawi), wshatanawi@psu.edu.sa (W. Shatanawi)

http://www.ejpam.com 702 c© 2018 EJPAM All rights reserved.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 703

Definition 1. [19] Let X be a nonempty set. A function p : X ×X → [0,∞) is called a
partial metric space if for all x, y, z ∈ X, the following conditions are satisfied:

(p1) x = y ⇔ p(x, x) = p(x, y) = p(y, y),

(p2) p(x, x) ≤ p(x, y),

(p3) p(x, y) = p(y, x),

(p4) p(x, y) ≤ p(x, z) + p(z, y)− p(z, z).

The pair (X, p) is called the notion of a partial metric space(PMS). The sequence {xn}
in X converges to a point x ∈ X if limn→∞ p(xn, x) = p(x, x). Also the sequence {xn}
is called p−Cauchy if the limn,m→∞p(xn, ym) exists. The partial metric space (X, p) is
called complete if for every p-Cauchy sequence {xn}n∞, there is some x ∈ X such that

p(x, x) = lim
n→∞

p(xn, x) = lim
n,m→∞

p(xn, xm).

A basic example of a partial metric space is the pair (R+, p), where p(x, y) = max{x, y}
for all x, y ∈ R+.

Harandi [14] introduced a new generalization of partial metric space, called a metric-
like space. He established the existence and uniqueness of fixed points in a metric-like
space as well as in a partially ordered metric-like space.

Definition 2. [14] Let X be a nonempty set. A function σ : X ×X → [0,∞) is said to
be a metric like space on X if for any x, y, z ∈ X, the following conditions hold:

(σ1) σ(x, y) = 0⇒ x = y,

(σ2) σ(x, y) = σ(y, x),

(σ3) σ(x, z) ≤ σ(x, y) + σ(y, z).

The pair (X,σ) is called a metric-like space.

It is clear that every partial metric space is a metric-like space but the converse is not
true.

Example 1. [14] Let X = {0, 1} and

σ(x, y) =


2, if x = y = 0;

1, otherwise.

Then (X,σ) is a metric-like space but it is not a partial metric space. Note that σ(0, 0) 6≤
σ(0, 1).



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 704

Moreover, each metric-like space σ on X generates a topology τσ on X whose base is
the family of open σ-balls

Bσ(x, ε) = {y ∈ X :| σ(x, y)− σ(x, x) |< ε}, for all x ∈ X and ε > 0.

Let (X,σ) and (Y, σ) be metric-like spaces, and let f : X → Y be a continuous mapping.
Then

lim
n→∞

xn = x ⇒ lim
n→∞

fxn = fx.

A sequence {xn}n=0
∞ of elements of X is called σ-Cauchy if the limit limn,m→∞ σ(xn, xm)

exists. The metric-like space (X,σ) is called complete if for each σ-Cauchy sequence
{xn}∞n , there exists x ∈ X such that

lim
n→∞

σ(xn, x) = σ(x, x) = lim
n,m→∞

σ(xn, xm).

Remark 1. [16] Let X = {0, 1}, and σ(x, y) = 1 for each x, y ∈ X. Consider the sequence
{xn} such that xn = 1 for each n ∈ N. Then it is easy to see that xn → 0 and xn → 1,
therefore the limit of a convergent sequence is not necessarily unique.

Lemma 1. [16] Let (X,σ) be a metric-like space. Let {xn} be a sequence in X that
converges to x ∈ X such that, σ(x, x) = 0. Then, for all y ∈ X, we have limn→∞ σ(xn, y) =
σ(x, y).

Example 2. Let X = R and σ : X ×X → [0,+∞) be defined by

σ(x, y) =


2k, if x = y = 0;

k, otherwise.

Then (X,σ) is a metric-like space, but for k > 0, it is not a partial metric space, as
σ(0, 0) 6≤ σ(0, 1).

Now let F be the family of all functions β : [0,∞)→ [0, 1) which satisfy the condition

limn→∞ β(tn) = 1 implies limn→∞ tn = 0.

In 2015, Karapinar et al.[18] proved the following particular result(it corresponds to
S = 1 and ψ(t) = t).

Theorem 1. [18] Let (X,σ) be a complete metric-like space and f : X → X be a mapping.
Suppose that there exists β ∈ F such that

σ(fx, fy) ≤ β(σ(x, y))σ(x, y), (1)

for all x, y ∈ X. Then f has a unique fixed point.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 705

In 2012, Samet et al. [26] introduced the concept of α-admissible mappings as the
following.

Definition 3. [26] Let f : X → X and α : X×X → [0,∞). Then f is called α-admissible
if for all x, y ∈ X with α(x, y) ≥ 1 implies α(fx, fy) ≥ 1.

Sintunavarat [30] presented the notion of weak α-admissible mappings as follows:

Definition 4. [30] Let X be a nonempty set and let α : X × X → [0,∞) be a given
mapping. A mapping f : X → X is said to be a weak α-admissible mappings if the
following condition holds:

x ∈ X with α(x, fx) ≥ 1⇒ α(fx, f2x) ≥ 1.

Remark 2. [30] It is customary to write A(X,α) and WA(X,α) to denote the collection
of all α-admissible mappings on X and the collection of all weak α-admissible mappings
on X. One can verify that A(X,α) ⊆ WA(X,α).

On the other hand, the concept of F -contraction was introduced by Wardowski in [31].

Definition 5. [31] Let F : R+ → R be a mapping satisfying the following:

(F1) F be a strictly increasing, that is, for α, β ∈ R+ such that α < β implies F (α) <
F (β),

(F2) For each sequence {αn} of positive numbers,
limn→∞ αn = 0 if and only if limn→∞ F (αn) = −∞,

(F3) There exists k ∈ (0, 1) such that limα→0+ α
kF (α) = 0.

Recently, Piri and Kumam [21] investigated some fixed point theorems concerning F -
contraction in complete metric spaces by replacing the condition (F3) with the condition:
(F3̀) F is continuous on (0,∞).

Definition 6. [31] Let (X, d) be a metric space. A mapping T : X → X is said to be an
F -contraction if there exist F ∈ F and τ > 0 such that

d(Tx, Ty) > 0⇒ (τ + F (d(Tx, Ty)) ≤ F (d(x, y)),

for all x, y ∈ X.

2. Main Result

In this section, we shall state and prove our main results. We firstly recall the following
classes of functions. Let F : R+ → R is strictly increasing contraction function. Let F be
the family of all functions β : [0,∞)→ [0, 1) which satisfy the condition

limn→∞ β(tn) = 1 implies limn→∞ tn = 0.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 706

Definition 7. Let (X,σ) be a metric-like space and α : X × X → [0,∞). A mapping
f : X → X is said to be an (α, β, F )-Geraghty contraction mapping if there exist β ∈ F
and τ > 0 such that, for all x, y ∈ X with σ(fx, fy) > 0 and α(x, y) ≥ 1,

α(x, y)(τ + F (σ(fx, fy)) ≤ β(Mx,y)F (Mx,y), (2)

where

Mx,y = max{σ(x, y), σ(x, fx), σ(y, fy),
σ(fx, y) + σ(x, fy)

4
,
[1 + σ(x, fx)]σ(y, fy)

σ(x, y) + 1
}.

Remark 3. Since the functions belonging to F are strictly smaller than 1, the expression
β(Mx,y) in 20 can be estimated from above as follows:

β(Mx,y) < 1,

for all x, y ∈ X with σ(fx, fy) > 0.

Lemma 2. Let (X,σ) be a metric-like space, and let f : X → X is said to be an
(α, β, F )−Geraghty contraction mapping. Define a sequence {xn} by xn+1 = fxn for
all n ∈ N. If the sequence {xn} is non-decreasing and limn→∞ σ(xn, xn+1) = 0, then {xn}
is a Cauchy sequence.

Proof. Suppose that the sequence {xn} is not a Cauchy, then there exists ε > 0 and two
subsequences {xpn} and {xqn} of the sequence {xn} such that pn > qn > n, σ(xpn−1, xqn) <
ε and σ(xpn , xqn) ≤ ε. This implies that

ε ≤ σ(xpn , xqn)

≤ σ(xpn , xqn−1) + σ(xqn−1, xqn)

≤ σ(xpn , xpn−1) + σ(xpn−1, xqn−1) + σ(xqn−1, xqn)

≤ σ(xpn , xpn−1) + σ(xpn−1, xqn) + 2σ(xqn−1, xqn)

< σ(xpn , xpn−1) + ε+ 2σ(xqn−1, xqn).

Since σ(xn, xn+1) 6= 0, we have

lim
n→∞

σ(xpn , xqn) = lim
n→∞

σ(xpn , xqn−1) (3)

= lim
n→∞

σ(xpn−1, xqn−1) (4)

= lim
n→∞

σ(xpn−1, xqn) (5)

= ε.

Since f is an (α, β, F )-Geraghty contraction mapping and α(x, y) ≥ 1, we have

(τ + F (σ(xpn−1, xqn−1))) ≤ α(xpn−1, xqn−1)(τ + F (σ(xpn−1, xqn−1))) ≤ β(Mxpn−1,xqn−1)F (Mxpn−1,xqn−1),

where

Mxpn−1,xqn−1 = max{σ(xpn−1, xqn−1), σ(xpn−1, fxpn−1), σ(xqn−1 , fxqn−1),



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 707

σ(fxpn−1, xqn−1) + σ(xpn−1, fxqn−1)

4
,

[1 + σ(xpn−1, fxpn−1)]σ(xqn−1, fxqn−1)

σ(xpn−1, xqn−1) + 1
}

= max{σ(xpn−1, xqn−1), σ(xpn−1, xpn), σ(xqn−1 , xqn),

σ(xpn , xqn−1) + σ(xpn−1, xqn)

4
,

[1 + σ(xpn−1, xpn)]σ(xqn−1, xqn)

σ(xpn−1, xqn−1) + 1
}.

Letting n→∞ in the above inequalities and using (2.2), (2.3) and (2.4), we obtain

lim
n→∞

M(xpn−1, xqn−1) = ε. (6)

Since limn→∞ β(M(xpn−1, xqn−1) ≤ 1, we conclude that

τ + F (ε) ≤ β(ε)F (ε)

≤ F (ε), (7)

a contradiction since τ > 0. Hence

lim
n→∞

σ(xn, xm) = 0.

We denote with Ξ(X,α, β, F ) the collection of all almost generalized (α, β, F )−contractive
mappings.

Theorem 2. Let (X,σ) be a metric-like space and α : X × X → [0,∞). A mapping
f : X → X be an (α, β, F )−Geraghty contraction mapping. Assume that the following
conditions are satisfied:

(i) f ∈ Ξ(X,α, β, F ) ∩WA(X,α).

(ii) There exists x0 ∈ X such that σ(x0, fx0) ≥ 1.

(iii) f is σ−continuous.

Then f has a unique fixed point z ∈ X with σ(z, z) = 0.

Proof. Let x0 ∈ X such that α(x0, fx0) ≥ 1. We define a sequence {xn} in X such that
xn = fxn−1 for all n ∈ N. If σ(xn, xn+1) = 0 for some n0 ∈ N, then xn0 is a fixed point of
f and it is done. Now, suppose that xn 6= xn+1 for all n ∈ N. Since f ∈ WA(X,α, β) and
α(x0, fx0) ≥ 1, we have

α(x1, x2) = α(fx0, ffx0) ≥ 1, α(x2, x3) = α(fx1, ffx1) ≥ 1.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 708

Using this process again, we get α(xn, xn+1) ≥ 1.
Since f : X → X is (α, β, F )-Geraghty contraction mapping with α(fxn−1, ffxn−1) =
α(xn, xn+1) ≥ 1, we have

0 < τ + F (σ(xn, xn+1)) ≤ α(xn, xn+1)(τ + F (σ(fxn−1, fxn))

≤ β(Mxn−1,xn)F (Mxn−1,xn), (8)

where

Mxn−1,xn = max{σ(xn−1, xn), σ(xn−1, fxn−1), σ(xn, fxn),

σ(xn−1, fxn) + σ(fxn−1, xn
4

,
[1 + σ(xn−1, fxn−1)]σ(xn, fxn)

σ(xn−1, xn) + 1
}

= max{σ(xn−1, xn), σ(xn−1, xn), σ(xn, xn+1),

σ(xn−1, xn+1) + σ(xn, xn
4

,
[1 + σ(xn−1, xn)]σ(xn, xn+1)

σ(xn−1, xn) + 1
}

= max{σ(xn−1, xn), σ(xn, xn+1),
σ(xn−1, xn+1)

4
, σ(xn, xn+1)}

< max{σ(xn−1, xn), σ(xn, xn+1),
σ(xn−1, xn) + σ(xn, xn+1)

4
}

= max{σ(xn−1, xn), σ(xn, xn+1),
σ(xn−1, xn) + σ(xn, xn+1)

4
}

= max{σ(xn−1, xn), σ(xn, xn+1)}. (9)

If max{σ(xn−1, xn), σ(xn, xn+1)} = σ(xn−1, xn), then

F (σ(xn−1, xn)) ≤ β(σ(xn−1, xn))F (σ(xn−1, xn)− τ
≤ F (σ(xn−1, xn)),

which is a contradiction. Thus,we conclude that

max{σ(xn−1, xn), σ(xn, xn+1)} = σ(xn, xn+1)

, for all n ∈ N. Then
F (σ(xn, xn+1)) ≤ F (σ(xn, xn+1))− τ

, for all n ∈ N. Repeating this process, we obtain

F (σ(xn, xn+1)) ≤ F (σ(x0, x1))− nτ (10)

By taking n→∞ in (2.11) that shows limn→∞ F (σ(xn, xn+1)) = −∞, hence

lim
n→∞

σ(xn, xn+1) = 0. (11)

Now, by Lemma 2, {xn} is a Cauchy sequence. Since X is complete, there exists z ∈ X
such that

lim
n→∞

σ(xn, z) = σ(z, z) = lim
n,m→∞

σ(xn, xm) = 0. (12)



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 709

Since f is continuous, we claim z = fz. Assume the contrary, that is z 6= fz. In this case,
there exists a sequence {xn} for n0 ∈ N such that σ(fxn, fz) > 0 for all n ≥ n0. Then
from our assumption (with n ≥ n0), we have

τ + F (σ(xn+1, fz)) = τ + F (σ(fxn, fz))

≤ α(xn, z)(τ + F (σ(xn, fz)))

≤ β(Mxn,z)F (Mxn,z), (13)

where

Mxn,z = max{σ(xn, z), σ(xn, fxn), σ(z, fz),
σ(fxn, z) + σ(xn, fz)

4
,

[1 + σ(xn, fxn)]σ(z, fz)

σ(xn, z) + 1
}

= max{σ(xn, z), σ(xn, xn+1), σ(z, fz),
σ(xn+1, z) + σ(xn, fz)

4
,

[1 + σ(xn, xn+1)]σ(z, fz)

σ(xn, z) + 1
}. (14)

By taking n→∞, we get

lim
n→∞

Mxn,z = max{σ(z, z), σ(z, fz), σ(z, fz),
σ(fz, z) + σ(z, fz)

4
,

[1 + σ(z, fz)]σ(z, fz)

σ(z, z) + 1
}

= max{σ(z, fz),
σ(z, fz)

4
}

= σ(z, fz). (15)

Therefore, by taking the limits as n→∞ in (2.12), we get

F (σ(z, fz)) ≤ β(σ(z, fz)))F (σ(z, fz))− τ
≤ F (σ(z, fz))− τ, (16)

which gives a contradiction. Hence, we conclude z is a fixed point of f .
Further, suppose that z, ź are two fixed points of f such that z 6= ź and α(fz, ff ź) =
α(z, ź) ≥ 1 and σ(fz, f ź) = σ(z, ź) ≥ 0. From (2.1), we have

τ + F (σ(z, ź)) = τ + F (σ(fz, f ź))

≤ α(z, ź)(τ + F (σ(fz, f ź)))

≤ β(Mz,ź)F (Mz,ź),

where

Mz,ź = max{σ(z, ź), σ(z, f ź), σ(ź, f ź),
σ(fz, ź) + σ(z, f ź)

4
,



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 710

[1 + σ(z, fz)]σ(ź, f ź)

σ(z, ź) + 1
}

= max{σ(z, ź), σ(z, ź), σ(ź, ź),
σ(z, ź)

2
, σ(ź, f ź)

= max{σ(z, ź),
σ(z, ź)

2
= σ(z, ź).

Hence

τ + F (σ(z, ź)) ≤ β(σ(z, ź))F (σ(z, ź))

≤ F (σ(z, ź)),

which is a contradiction. Hence σ(z, ź) = 0, that is z = ź. Thus, we conclude that the
fixed point of f is unique.

Next, we will prove that σ(z, z) = 0. If σ(fz, fz) = σ(z, z) > 0 and α(fz, ffz) =
α(z, z) ≥ 1, then from (2.1)and applying the routine calculation as mentioned above, we
get

τ + F (σ(z, z)) = τ + F (σ(fz, fz))

≤ α(z, z)(τ + F (σ(fz, fz)))

≤ β(Mz,z)F (Mz,z),

where

Mz,z = max{σ(z, z), σ(z, fz), σ(z, fz),
σ(fz, z) + σ(z, fz)

4
,
[1 + σ(z, fz)]σ(z, fz)

σ(z, z) + 1
}

= σ(z, z).

Hence

τ + F (σ(z, z)) < β(σ(z, z))F (σ(z, z))

≤ F (σ(z, z)),

is a contradiction, thus, σ(z, z) = 0.

The following two corollaries are direct results of Theorem 2.

Corollary 1. Let (X,σ) be a complete metric-like space, α : X × X → [0,∞) and f :
X → X be two given mapping satisfying the following conditions:

(i) f ∈ Ξ(X,α, β, F ) ∩WA(X,α).

(ii) There exists x0 ∈ X such that σ(x0, fx0) ≥ 1.

(iii) f is σ−continuous.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 711

Then f has a unique fixed point z ∈ X such that σ(z, z) = 0.
Proof. It follows from Theorem 2 by putting

Mx,y = max{σ(x, y), σ(x, fx), σ(y, fy)}.

Corollary 2. Let (X,σ) be a complete metric-like space, α : X × X → [0,∞) and f :
X → X be two given mapping satisfying the following conditions:

(i) f ∈ Ξ(X,α, β, F ) ∩WA(X,α).

(ii) There exists x0 ∈ X such that σ(x0, fx0) ≥ 1.

(iii) f is σ−continuous.

Then f has a unique fixed point z ∈ X such that σ(z, z) = 0.
Proof. It follows from Theorem 2 by putting Mx,y = aσ(x, y) + bσ(x, fx) + cσ(y, fy) +

e[σ(fx,y)+σ(x,fy)
4 ] + e[ [1+σ(x,fx)]σ(y,fy)

σ(x,y)+1 ].
For all x, y ∈ X, we have

Mx,y = aσ(x, y) + bσ(x, fx) + cσ(y, fy) + e[
σ(fx, y) + σ(x, fy)

4
]

≤ (a+ b+ c+ 2e) max{σ(x, y), σ(x, fx), σ(y, fy),
σ(fx, y) + σ(x, fy)

4
,
[1 + σ(x, fx)]σ(y, fy)

σ(x, y) + 1

[1 + σ(x, fx)]σ(y, fy)

σ(x, y) + 1
}

≤ max{σ(x, y), σ(x, fx), σ(y, fy),
σ(fx, y) + σ(x, fy)

4
,
[1 + σ(x, fx)]σ(y, fy)

σ(x, y) + 1
}.

Then, we see that (2.1) is a consequence of (2.14), then the corollary is proved.

Example 3. Let X = {0, 1, 2}. Let σ : X ×X → R be a metric like function define by
σ(0, 0) = σ(1, 1) = σ(2, 2) = 0,
σ(1, 2) = σ(2, 1) = 3,
σ(2, 0) = σ(0, 2) = 2,
σ(0, 1) = σ(1, 0) = 3

2 .
It is easy to see that (X,σ) is a complete metric-like space. Also, define f : X → X be
given by f0 = 0 = f1 and f2 = 1.

Define α : [0,+∞)→ [0, 1) by

α(x, y) =

{
1 if x ∈ {0, 1, 2}
0 if otherwise.

Define β : [0,∞)→ [0, 1) by

β(t) =


1

1 + 1
7 t

if t > 0

1

2
if t = 0.



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 712

Suppose that F (t) = et and τ = 1
4 . The function (f) satisfies the inequality (20). For

that, given x, y ∈ X. Then we have the following cases:

Case 1: x = 0 and x = 1.
Then α(0, 1) = 1 and M0,1 = max{0, 0, 0, 0, 1} = 1.
σ(f0, f1) = σ(0, 0) = 0. Now

0 < α(0, 1)(τ + F (σ(f0, f1))) = τ + F (σ(0, 0))

= (τ + F (0)

= τ

≤ β(M0,1)F (M0,1)

= β(1)F (1)

= e (17)

Case 2: x = 0 and y = 2.
Then α(0, 2) = 1 and M0,2 = max{2, 0, 3, 13

16 , 1} = 3.
σ(f0, f2) = σ(0, 1) = 3

2 . Now

0 < α(0, 2)(τ + F (σ(f0, f2))) = τ + F (
3

2
)

= τ +
3

2
≤ β(M0,2)F (M0,2)

= β(3)F (3)

= 3e3,

Case 3: x = 1 and y = 2.
Then α(1, 2) = 1 and M1,2 = max{3, 3

2 , 3,
1
2 ,

15
8 } = 3.

σ(f1, f2) = σ(0, 1) = 3
2 . Now

0 < α(0, 2)(τ + F (σ(f0, f2))) = τ + F (
3

2
)

= τ +
3

2
≤ β(M0,2)F (M0,2)

= β(3)F (3)

= 3e3,

Thus, all the conditions of Theorem 2 are satisfied and hence f has a unique fixed
point.

3. Consequences

In this section, we derive the analog of Theorem 2 in the context of partial metric
spaces (PMS). In the following theorem we conclude the existence and the uniqueness of



H. Qawaqneh, M.S. Noorani, W. Shatanawi / Eur. J. Pure Appl. Math, 11 (3) (2018), 702-716 713

a fixed point of the given mapping.

Theorem 3. Let (X, p) be a a complete partial metric space and α : X ×X → [0,∞). A
mapping f : X → X be an (α, β, F )−Geraghty contraction mapping. Suppose there exist
f ∈ F and τ > 0 such that, for all x, y ∈ X with σ(fx, fy) > 0 and α(x, y) ≥ 1,

0 < α(x, y)(τ + F (σ(fx, fy)) ≤ β(Mx,y)F (Mx,y), (18)

where

Mx,y = max{max{p(x, y), p(x, fx), p(y, fy),
p(fx, y) + p(x, fy)

4
,
[1 + p(x, fx)]p(y, fy)

p(x, y) + 1
}.

Then f has a unique fixed point z ∈ X with p(z, z) = 0.
Proof. Since every partial metric space is a metric-like space, we obtain the proof by

following the proof in Theorem 2. We now show the uniqueness of the fixed point of f.
Suppose there is another fixed point y∗ ∈ X of f , such that x∗ 6= y∗. Thus from Lemma
??, we have p(x∗, y∗) > 0. From (p2), we have p(fx∗, fy∗) = p(x∗, y∗) > 0. Thus

0 < τ + F (p(x∗, y∗)) ≤ α(x∗, y∗)(τ + F (p(fx∗, fy∗))

≤ β(Mx∗,y∗)F (Mx∗,y∗)

= β(p(x∗, y∗))F (p(x∗, y∗))

≤ F (p(x∗, y∗)),

where

Mx∗,y∗ = max{p(x∗, y∗), p(x∗, fx∗), p(y∗, fy∗), p(fx
∗, y∗) + p(x∗, fy∗

4
,

[1 + p(x∗, fx∗)]p(y∗, fy∗)

p(x∗, y∗) + 1
}

= max{p(x∗, y∗), p(x∗, x∗), p(y∗, y∗), p(x
∗, y∗) + p(x∗, y∗)

4
,

[1 + p(x∗, y∗)]p(y∗, y∗)

p(x∗, y∗) + 1
}

= max{p(x∗, y∗), p(x∗, x∗), p(y∗, y∗), p(x
∗, y∗)

2
, p(y∗, y∗)}

= p(x∗, y∗).

This is a contradiction, and hence x∗ = y∗.

Theorem 4. Let (X, p) be a a complete partial metric space and α : X ×X → [0,∞). A
mapping f : X → X be an (α, β, F )−Geraghty contraction mapping. Suppose there exist
f ∈ F and τ > 0 such that, for all x, y ∈ X with σ(fx, fy) > 0 and α(x, y) ≥ 1,

0 < α(x, y)(τ + F (σ(fx, fy)) ≤ β(Mx,y)F (Mx,y), (19)

where
Mx,y = max{max{p(x, y), p(x, fx), p(y, fy)}.

Then f has a unique fixed point z ∈ X with p(z, z) = 0.



REFERENCES 714

Theorem 5. Let (X, p) be a a complete partial metric space and α : X × X → [0,∞).
A mapping f : X → X be an (α, β, F )Geraghty contraction mapping. Suppose there exist
f ∈ F and τ > 0 such that, for all x, y ∈ X with σ(fx, fy) > 0 and α(x, y) ≥ 1,

0 < α(x, y)(τ + F (σ(fx, fy)) ≤ β(Mx,y)F (Mx,y), (20)

where
Mx,y = p(x, y).

Then f has a unique fixed point z ∈ X with p(z, z) = 0.

Acknowledgements

The authors would like to acknowledge the grant: UKM Grant DIP-2017-011 and
Ministry of Education, Malaysia grant FRGS/1/2017/STG06/UKM/01/1 for financial
support.

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