International Journal of Analysis and Applications
ISSN 2291-8639
Volume 5, Number 1 (2014), 91-101
http://www.etamaths.com

FIXED POINTS UNDER ψ-α-β CONDITIONS IN ORDERED

PARTIAL METRIC SPACES

ZORAN KADELBURG1,∗ AND STOJAN RADENOVIĆ2

Abstract. Recently, E. Karapinar and P. Salimi [Fixed point theorems via

auxiliary functions, J. Appl. Math. 2012, Article ID 792174] have obtained
fixed point results for increasing mappings in a partially ordered metric space

using three auxiliary functions in the contractive condition. In this paper,

these results are extended to 0-complete ordered partial metric spaces with a
more general contractive condition. Examples are given showing that these

extensions are proper.

1. Introduction

Matthews [1] introduced the notion of a partial metric space as a part of the
study of denotational semantics of dataflow networks. He showed that the Banach
contraction mapping theorem can be generalized to the partial metric context for
applications in program verification. Subsequently, several authors (see, e.g., [2, 3,
4, 5, 6]) proved variuos more general fixed point results in partial metric spaces.

The notion of weakly contractive conditions in metric spaces was first used by
Rhoades [7] who proved the following

Theorem 1.1. [7] Let (X,d) be a metric space. If T : X → X satisfies the
condition

d(Tx,Ty) ≤ d(x,y) −ϕ(d(x,y)), ∀x,y ∈ X,

where ϕ : [0, +∞) → [0, +∞) is a continuous and nondecreasing function such that
ϕ(t) = 0 iff t = 0, then T has a unique fixed point.

Subsequently, several authors (see, e.g., [8, 9]) proved various generalizations and
refinements of this result.

Fixed point theory has developed rapidly in partially ordered metric spaces, that
is, metric spaces endowed with a partial ordering. The first result in this direction
was given by Ran and Reurings [10] who presented its applications to matrix e-
quations. Further, a lot of results appeared, we mention just those contained in
[11, 12, 13, 14, 15, 16]. These results use weaker contractive conditions (mostly just
for comparable elements of the given space), but at the expense of some additional
restrictions to the mappings involved.

2010 Mathematics Subject Classification. 47H10, 54H25.
Key words and phrases. Common fixed point, partial metric space, weakly contractive

condition.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

91



92 KADELBURG AND RADENOVIĆ

Consider the following classes of functions from [0, +∞) into itself:

Ψ = {ψ : ψ is nondecreasing and lower semicontinuous },
Φ1 = {α : α is upper semicontinuous },
Φ2 = {β : β is lower semicontinuous }.

Very recently, Karapinar and Salimi [16] proved the following

Theorem 1.2. Let (X,�,d) be a complete ordered metric space and let T : X → X
be a nondecreasing selfmap. Assume that there exist ψ ∈ Ψ, α ∈ Φ1 and β ∈ Φ2
such that for all s,t ≥ 0,

(1.1) t > 0 and (s = t or s = 0) implies ψ(t) −α(s) + β(s) > 0,

and

(1.2) ψ(d(Tx,Ty)) ≤ α(d(x,y)) −β(d(x,y))

for all comparable x,y ∈ X. Suppose that, either, T is continuous, or X is regular
(see further Definition 2.5). If there exists x0 ∈ X such that x0 � Tx0, then T has
a fixed point.

We shall prove in the present paper that this result can be extended to 0-complete
ordered partial metric spaces with a contractive condition more general than (1.2).
Examples will be given showing that this extension is proper.

Remark 1.3. It was shown very recently (see [17]) that in some cases fixed point
results in partial metric spaces can be directly reduced to their standard metric
counterparts. We note that the results of the present paper do not fall into this
category. Moreover, using the method from [17], it is not possible to conclude that
if x is a fixed point of the mapping under consideration, then p(x,x) = 0, which is
an important conclusion for applications in Computer Science.

2. Preliminaries and auxiliary results

The following definitions and details can be seen in [1, 2, 3, 4, 18].

Definition 2.1. A partial metric on a nonempty set X is a function p : X×X → R+
such that for all x,y,z ∈ X:

(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).

A partial metric space is a pair (X,p) of a nonempty set X and a partial metric p
on X. If, moreover, � is a partial order on X, then the triple (X,�,p) is called an
ordered partial metric space.

It is clear that, if p(x,y) = 0, then from (p1) and (p2), x = y. But p(x,x) may not
be 0. A sequence {xn} in (X,p) converges to a point x ∈ X if limn→∞p(x,xn) =
p(x,x). This will be denoted as xn → x (n → ∞) or limn→∞xn = x. Clearly, a
limit of a sequence in a partial metric space need not be unique. Moreover, the
function p(·, ·) need not be continuous in the sense that xn → x and yn → y imply
p(xn,yn) → p(x,y).



FIXED POINTS IN PARTIAL METRIC SPACES 93

Example 2.2. (1) A paradigmatic example of a partial metric space is the pair
(R+,p), where p(x,y) = max{x,y} for all x,y ∈ R+.

(2) [1] Let X = { [a,b] : a,b ∈ R, a ≤ b} and let p([a,b], [c,d]) = max{b,d}−
min{a,c}. Then (X,p) is a partial metric space.

Definition 2.3. Let (X,p) be a partial metric space. Then:

(1) A sequence {xn} in (X,p) is called a Cauchy sequence if limn,m→∞p(xn,xm)
exists (and is finite).

(2) The space (X,p) is said to be complete if every Cauchy sequence {xn} in
X converges to a point x ∈ X such that p(x,x) = limn,m→∞p(xn,xm).

(3) [2] A sequence {xn} in (X,p) is called 0-Cauchy if limn,m→∞p(xn,xm) = 0.
The space (X,p) is said to be 0-complete if every 0-Cauchy sequence in X
converges (in τp) to a point x ∈ X such that p(x,x) = 0.

Lemma 2.4. Let (X,p) be a partial metric space.
(a) [3] If p(xn,z) → p(z,z) = 0 as n → ∞, then p(xn,y) → p(z,y) as n → ∞

for each y ∈ X.
(b) [2] If (X,p) is complete, then it is 0-complete.

The converse assertion of (b) does not hold as an easy example in [2] shows.

Definition 2.5. Let (X,�,p) be an ordered partial metric space. We say that X is
regular if the following holds: if {zn} is a nondecreasing sequence in X with respect
to � such that zn → z ∈ X as n →∞ (in (X,p)), then zn � z for all n ∈ N and if
{zn} is a nonincreasing sequence in X with respect to � such that zn → z ∈ X as
n →∞ (in (X,p)), then zn � z for all n ∈ N.

We will also need the following lemma which was proved in the metric case, e.g.,
in [19]. The proof is similar in the partial metric case, and so we omit it.

Lemma 2.6. Let (X,p) be a partial metric space and let {xn} be a sequence in X
such that

(2.1) lim
n→∞

p(xn+1,xn) = 0.

If {xn} is not a 0-Cauchy sequence in (X,p), then there exist ε > 0 and two
sequences {mk} and {nk} of positive integers such that nk > mk > k and the
following four sequences tend to ε+ when k →∞:

(2.2) p(xmk,xnk ), p(xmk,xnk+1), p(xmk−1,xnk ), p(xmk−1,xnk+1).

3. Main results

Our first result is the following

Theorem 3.1. Let (X,�,p) be a 0-complete ordered partial metric space and let
T : X → X be a nondecreasing selfmap. Assume that there exist functions ψ ∈ Ψ,
α ∈ Φ1 and β ∈ Φ2 such that for all t,s ≥ 0,

(3.1) t > 0 and (s = t or s = 0) implies ψ(t) −α(s) + β(s) > 0,

and

(3.2) ψ(p(Tx,Ty)) ≤ α(M(x,y)) −β(M(x,y))



94 KADELBURG AND RADENOVIĆ

for all comparable x,y ∈ X, where

M(x,y) = max

{
p(x,y),p(x,Tx),p(y,Ty),

1

2
[p(x,Ty) + p(y,Tx)]

}
.

Suppose that, either T is continuous, or X is regular. If there exists x0 ∈ X such
that x0 � Tx0, then T has a fixed point z ∈ X satisfying that p(z,z) = 0.

Proof. Starting with the given x0, construct the Picard sequence {xn} by xn+1 =
Txn, n ∈ N0. Using the assumption x0 � Tx0 and that T is nondecreasing, we
conclude that

x0 � x1 � ···� xn � xn+1 � ··· .
If xn+1 = xn for some n ∈ N0, a fixed point of T is found. Suppose that xn+1 6= xn
for all n ∈ N0. Apply assumption (3.2) for x = xn and y = xn+1 to obtain

ψ(p(xn+1,xn+2)) = ψ(p(Txn,Txn+1))(3.3)

≤ α(M(xn,xn+1)) −β(M(xn,xn+1)),

where

M(xn,xn+1) = max{p(xn,xn+1),p(xn,xn+1),p(xn+1,xn+2),
1
2
(p(xn,xn+2) + p(x2n+1,x2n+1))}

= max{p(xn,xn+1),p(xn+1,xn+2)}

(condition (p4) of partial metric was used). Suppose that p(xn+1,xn+2) > p(xn,xn+1)
for some n ∈ N0. Then (3.3) implies that

ψ(p(xn+1,xn+2)) ≤ α(p(xn+1,xn+2)) −β(p(xn+1,xn+2)).

By the assumption (3.1) it follows that p(xn+1,xn+2) = 0 and, hence, xn+1 = xn+2,
which is already excluded. Hence,

p(xn+1,xn+2) ≤ p(xn,xn+1),

and M(xn,xn+1) = p(xn,xn+1) for all n ∈ N0. Thus, the sequence {p(xn,xn+1)}
is nonincreasing. Since it is bounded from below, there exists r ≥ 0 such that
limn→∞p(xn,xn+1) = r. It follows from (3.3) that

ψ(p(xn+1,xn+2) ≤ α(p(xn,xn+1)) −β(p(xn,xn+1)),

and using the properties of functions ψ,α,β we get that

ψ(r) ≤ lim inf ψ(p(xn+1,xn+2)) ≤ lim sup ψ(p(xn+1,xn+2))
≤ lim sup[α(p(xn,xn+1)) −β(p(xn,xn+1))]
= lim sup α(p(xn,xn+1)) − lim inf β(p(xn,xn+1))]
≤ α(r) −β(r).

Using again the condition (3.1), we get that it is only possible if r = 0.
Next, we claim that {xn} is a 0-Cauchy sequence in the partial metric space

(X,p). Suppose that this is not the case. Then, using Lemma 2.6 we get that
there exist ε > 0 and two sequences {mk} and {nk} of positive integers such that
nk > mk > k and sequences (2.2) tend to ε when k →∞. Applying condition (3.2)
to elements x = xnk−1 and y = xmk we get that

(3.4) ψ(p(xnk,xmk+1)) ≤ α(M(xnk−1,xmk )) −β(M(xnk−1,xmk )),



FIXED POINTS IN PARTIAL METRIC SPACES 95

where

M(xnk−1,xmk ) = max{p(xnk−1,xmk ),p(xnk−1,xnk ),p(xmk,xmk+1),
1
2
[p(xnk−1,xmk+1) + p(xnk,xmk )]}.

Using Lemma 2.6 we get that limk→∞M(xnk−1,xmk ) = ε. Passing to the upper
limit in (3.4) and using properties of the functions ψ,α,β, we get that

ψ(ε) ≤ lim inf ψ(p(xnk,xmk+1)) ≤ lim sup ψ(p(xnk,xmk+1))
≤ lim sup[α(M(xnk−1,xmk )) −β(M(xnk−1,xmk ))]
= lim sup α(M(xnk−1,xmk )) − lim inf β(M(xnk−1,xmk ))
≤ α(ε) −β(ε).

This is (because of ε > 0) a contradiction with (3.1). We conclude that {xn} is a
0-Cauchy sequence in (X,p).

Now, since (X,p) is a 0-complete partial metric space, it follows that there exists
x ∈ X such that xn → x as n →∞, i.e.,

p(xn,x) → p(x,x) = 0, as n →∞.

Assume first that the mapping T is continuous. Then limn→∞p(Txn,Tx) =
p(Tx,Tx). It follows that

p(Tx,x) ≤ p(Tx,Txn) + p(xn+1,x) −p(xn+1,xn+1)
≤ p(Tx,Txn) + p(xn+1,x) → p(Tx,Tx) as n →∞.

It follows that p(Tx,x) = p(Tx,Tx). Since x � x, we can apply (3.2) to obtain

ψ(p(Tx,Tx)) ≤ α(M(x,x)) −β(M(x,x)) = α(p(x,Tx)) −β(p(x,Tx))
= α(p(Tx,Tx)) −β(p(Tx,Tx)).

By (3.1), this is possible only if p(Tx,Tx) = 0, i.e., p(x,Tx) = 0. Hence Tx = x.
Assume now that the space (X,�,p) is regular. Since the sequence {xn} is

increasing and xn → x as n → ∞, we get that xn � x for n ∈ N. Hence, we can
apply (3.2) to get

(3.5) ψ(p(Txn,Tx)) ≤ α(M(xn,x)) −β(M(xn,x)),

where

M(xn,x) = max{p(xn,x),p(xn,xn+1),p(x,Tx), 12 [p(xn,Tx) + p(xn+1,x)]}
→ p(x,Tx) as n →∞.

On the other hand, since p(xn,x) → p(x,x) = 0, Lemma 2.4.(1) implies that
p(Txn,Tx) = p(xn+1,Tx) → p(x,Tx). Hence, passing to the upper limit in (3.5),
similarly as in the previous case we get that it is only possible that p(x,Tx) = 0,
i.e., Tx = x.

In all possible cases we have obtained that x is a fixed point of the mapping T
satisfying p(x,x) = 0 and the theorem is proved. �

Remark 3.2. Taking p to be a standard metric in the contractive condition (3.2),
and assuming ψ = α to be continuous, we obtain [13, Corollary 3.3].

In a similar way one can prove the following two assertions.



96 KADELBURG AND RADENOVIĆ

Theorem 3.3. Let all the conditions of Theorem 3.1 be fulfilled, except that con-
dition (3.2) is replaced by

(3.6) ψ(p(Tx,Ty)) ≤ α(p(x,y)) −β(p(x,y)).

Then T has a fixed point in X.

Theorem 3.4. Let all the conditions of Theorem 3.1 be fulfilled, except that con-
dition (3.2) is replaced by

(3.7) ψ(p(Tx,Ty)) ≤ α(M1(x,y)) −β(M1(x,y)),

where

M1(x,y) = max

{
p(x,y),

1

2
[p(x,Tx) + p(y,Ty)],

1

2
[p(x,Ty) + p(y,Tx)]

}
.

Then T has a fixed point in X.

The following simple example shows that conditions of Theorem 3.1 are not
sufficient for the uniqueness of fixed points.

Example 3.5. Let X = {(1, 0), (0, 1)}, let (a,b) � (c,d) if and only if a ≤ c and b ≤
d, and let p be the Euclidean metric. The function T((a,b)) = (a,b) is continuous.
The only comparable pairs of points in X are x � x for x ∈ X and then M(x,x) = 0
and p(Tx,Tx) = 0, hence the condition ψ(p(fx,fy)) ≤ α(M(x,y))−β(M(x,y)) is
fulfilled, e.g., for the functions ψ ∈ Ψ, α ∈ Φ1, β ∈ Φ2 given as ψ(t) = α(t) = t,
β(t) = kt, 0 < k < 1. However, T has two fixed points (1, 0) and (0, 1).

Using the same example we can show that there exist situations where conditions
of Theorem 3.1, taken in the case without order may not be sufficient.

Example 3.6. If the previous example is considered without order, then one has
also to take into account the case when x 6= y. But then p(x,y) =

√
2 and

M(Tx,Ty) =
√

2, and so the condition (3.2) reduces to ψ(
√

2) ≤ α(
√

2) −β(
√

2)
and cannot be valid for any functions ψ,α,β satisfying (3.1).

Now we give a sufficient condition for the uniqueness of fixed point.

Theorem 3.7. Let all the conditions of Theorem 3.1 be fulfilled and, moreover, the
space (X,�,p) satisfy the following condition: For all x,y ∈ X there exists z ∈ X,
z � Tz, satisfying both x � z and y � z or there exists z ∈ X, z � Tz, satisfying
both x � z and y � z. Then the fixed point of T is unique.

Proof. Let x and y be two fixed points of T, i.e., Tx = x and Ty = y. Consider
the following two possible cases.

1. x and y are comparable. Then we can apply condition (3.2) and obtain that

ψ(p(x,y)) = ψ(p(Tx,Ty)) ≤ α(M(x,y)) −β(M(x,y)),

where

M(x,y) = max{p(x,y),p(x,Tx),p(y,Ty), 1
2
[p(x,Ty) + p(y,Tx)]}

= p(x,y)

and hence ψ(p(x,y)) ≤ α(p(x,y)) −β(p(x,y)) which is possible only if p(x,y) = 0
and hence x = y.



FIXED POINTS IN PARTIAL METRIC SPACES 97

2. Suppose now that x and y are not comparable. Choose an element z ∈ X,
z � Tz comparable with both of them. Then also x = Tnx is comparable with Tnz
for each n (since T is nondecreasing). Applying (3.1) one obtains that

ψ(p(x,Tnz)) = ψ(p(TTn−1x,TTn−1z))

≤ α(M(Tn−1x,Tn−1z)) −β(M(Tn−1x,Tn−1z)),

where

M(Tn−1x,Tn−1z) = max{p(Tn−1x,Tn−1z),p(Tn−1x,Tnx),p(Tn−1z,Tnz),
1
2
[p(Tn−1x,Tnz) + p(Tnx,Tn−1z)]}

= max{p(x,Tn−1z),p(Tn−1z,Tnz), 1
2
[p(x,Tnz) + p(x,Tn−1z))}

≤ max{p(x,Tn−1z),p(x,Tnz)},

for n sufficiently large, because p(Tn−1z,Tnz) → 0 when n →∞ (the last assertion
can be proved, starting from the assumption z � Tz, in the same way as a similar
conclusion in the proof of Theorem 3.1).

Similarly as in the proof of Theorem 3.1, it can be shown that p(x,Tnz) ≤
M(x,Tn−1z) ≤ p(x,Tn−1z). It follows that the sequence p(x,Tnz) is nonincreasing
and it has a limit l ≥ 0. Assuming that l > 0 and passing to the limit in the relation

ψ(p(x,Tnz)) ≤ α(M(x,Tn−1z)) −β(M(x,Tn−1z))

one obtains that l = 0, a contradiction. In the same way it can be deduced that
p(y,Tnz) → 0 as n → ∞. Now, passing to the limit in p(x,y) ≤ p(x,Tnz) +
p(Tnz,y), it follows that p(x,y) = 0. Hence, x = y and the uniqueness of the fixed
point is proved. �

Remark 3.8. If two selfmaps T,S : X → X are given, then in a similar way
a common fixed point result can be obtained, under an appropriate contractive
condition and assuming, e.g., that T and S are weakly increasing (for the details
see, e.g., [13]). The proof uses the standard method of Jungck sequences.

4. Examples

Our first example (inspired by [5]) shows that it may happen that the contrac-
tive condition (3.7) is not satisfied, hence the existence of a fixed point cannot be
obtained using Theorem 3.3. However, the condition (3.2) is fulfilled and Theorem
3.1 can be used to obtain the conclusion.

Example 4.1. Consider the set X = {a,b,c} and the function p : X × X → R
given by p(a,b) = p(b,c) = 1, p(a,c) = 3

2
, p(x,y) = p(y,x), p(a,a) = p(c,c) = 1

2
and p(b,b) = 0. Obviously, p is a partial metric on X, not being a metric (since
p(x,x) 6= 0 for x = a and x = c). Define an order-relation � on X by a � b � c.
Then, (X,�,p) is a 0-complete ordered partial metric space. Define a selfmap T
on X by

T :

(
a b c
b b a

)
.

Then T is not a (Banach)-contraction since

p(fc,fc) = p(a,a) =
1

2
= p(c,c)



98 KADELBURG AND RADENOVIĆ

and there is no λ ∈ [0, 1) such that p(fc,fc) ≤ λp(c,c). Moreover, condition (3.7)
(i.e., condition (1.2) of Theorem 1.2, with d replaced by p) cannot hold for functions
ψ ∈ Ψ, α ∈ Φ1, β ∈ Φ2 satisfying (1.1) because for otherwise x = y = c would
imply ψ( 1

2
) −α( 1

2
) + β( 1

2
) ≤ 0 which cannot hold.

We will check that T satisfies the condition (3.2) of Theorem 3.1 with functions
ψ,α,β given as ψ(t) = t, α(t) = t, β(t) = 1

3
t (obviously belonging to the respective

classes). Note that in this case α(t) −β(t) = 2
3
t. If x,y ∈{a,b}, then p(Tx,Ty) =

p(b,b) = 0 and (3.2) trivially holds. Let, e.g., y = c; then we have the following
three cases:

ψ(p(Ta,Tc)) = p(b,a) = 1 ≤
2

3
·

3

2

=
2

3
max{p(a,c),p(a,Ta),p(c,Tc), 1

2
[p(a,Tc) + p(c,Ta)]},

ψ(p(Tb,Tc)) = p(b,a) = 1 ≤
2

3
·

3

2

=
2

3
max{p(b,c),p(b,Tb),p(c,Tc), 1

2
[p(b,Tc) + p(c,Tb)]},

ψ(p(Tc,Tc)) = p(a,a) =
1

2
<

2

3
·

3

2
=

2

3
max{p(c,c),p(c,Tc)}.

Thus, conditions of Theorem 3.1 are satisfied and the existence of a fixed point of
T follows.

Using an example of Romaguera [2], we present another example showing how
Theorem 3.1 can be used. It also shows that there are situations when standard
metric arguments cannot be used to obtain the existence of a fixed point.

Example 4.2. Let X = [0, 1]∩Q be equipped with the partial metric p defined by
p(x,y) = max{x,y} for x,y ∈ X and the standard order. Let T : X → X be given
by

Tx =
x2

1 + x
.

It is easy to see that the space (X,p) is 0-complete. Take the functions ψ,α,β given
by ψ(t) = α(t) = t, β(t) = 1

2
t. The contractive condition (3.2) for (say) x ≥ y takes

the form

p(Tx,Ty) = max

{
x2

1 + x
,
y2

1 + y

}
=

x2

1 + x

≤
1

2
max{p(x,y),p(x, x

2

1+x
),p(y, y

2

1+y
), 1

2
[p(x, y

2

1+y
) + p(y, x

2

1+x
)]}

=
1

2
max{x,x,y, 1

2
[x + max{y, x

2

1+x
}]} =

1

2
x,

and it is satisfied for all x,y ∈ X, since 0 ≤ x ≤ 1. Hence, all the conditions of
Theorem 3.1 and Theorem 3.7 are satisfied and T has a unique fixed point (z = 0).

However, if we consider the same ordered set X = [0, 1] ∩ Q equipped with the
standard metric d(x,y) = |x−y|, we obtain a non-complete ordered metric space.
Hence, the existence of a fixed point of T cannot be deduced using Theorem 1.2.

We present now an example (inspired by [16]) showing the usage of Theorem 3.1
with (at least some of) functions ψ,α,β not being continuous.



FIXED POINTS IN PARTIAL METRIC SPACES 99

Example 4.3. Consider the set X = [0, 1] equipped with the standard order and
the partial metric given as p(x,y) = max{x,y}. It is easy to show that (X,�,p)
is a regular, 0-complete ordered partial metric space. Let T : X → X be given by
Tx = 1

2
x− 1

4
x2 and take the functions ψ,α,β defined by

ψ(t) =

{
t + 3

2
, t > 0

1, t = 0,
, α(t) = t +

5

2
, β(t) =

1

2
t + 1.

Note that ψ is lower semicontinuous and the condition (3.1) is satisfied.
Both conditions (3.2) and (3.7) are satisfied. We shall check, e.g., condition

(3.2). Let x,y ∈ X and, for example, x ≤ y. Then

ψ(p(Tx,Ty)) = ψ(Ty) =

{
1
2
y − 1

4
y2 + 3

2
, y > 0

1, y = 0,

and

M(x,y) = max{max{x,y}, max{x,Tx}, max{y,Ty}, 1
2
[max{x,Ty} + max{y,Tx}]}

= max{y,x,y, 1
2
[max{x,Ty} + y]} = y,

since max{x,Ty} ≤ y. The condition (3.2) reduces to 1
2
y − 1

4
y2 + 3

2
≤ 1

2
y + 3

2
if

y > 0 and to 1 ≤ 3
2

if y = 0 and it is fulfilled in both cases. By Theorems 3.1 and
3.7, T has a unique fixed point (which is z = 0).

Finally, the following example demonstrates the situation when contractive con-
ditions using partial metric can guarantee the existence of a fixed point, while the
respective conditions with the standard metric cannot.

Example 4.4. Consider the set X = [0, +∞) endowed with the partial metric
p(x,y) = max{x,y} and the order given by

x � y =⇒ x = y ∨ (x,y ∈ [0, 1] ∧x ≤ y).

(X,�,p) is a regular, 0-complete ordered partial metric space. Let T : X → X be
given by

Tx =




x2

1 + x
, x ∈ [0, 1]

x

2
, x > 1

and consider the functions ψ ∈ Ψ, α ∈ Φ1, β ∈ Φ2 defined as

ψ(t) = α(t) = t, β(t) =




t

1 + t
, t ∈ [0, 1]

t

2
, t > 1.

Clearly, ψ,α,β satisfy condition (3.1). We shall show that also condition (3.2)
is satisfied. Let x,y ∈ X and suppose that, e.g., y � x. Then, there are two
possibilities:

1. x ∈ [0, 1] (and hence also y ∈ [0, 1] and y ≤ x). Then

p(Tx,Ty) = max

{
x2

1 + x
,
y2

y + x

}
=

x2

1 + x
,



100 KADELBURG AND RADENOVIĆ

and

M(x,y) = max

{
x,x,y,

1

2

[
x + max

{
y,

x2

1 + x

}]}
= x,

since max{y, x
2

1+x
}≤ x. Condition (3.2) reduces to

x2

1 + x
≤ x−

x

1 + x

and obviously holds true.
2. x > 1 (and hence y = x). Then p(Tx,Ty) = x

2
and M(x,y) = x. Hence, (3.2)

reduces to x
2
≤ x− x

2
and is also satisfied.

All the conditions of Theorems 3.1 and 3.7 are fulfilled and T has a unique fixed
point (which is z = 0).

Consider now the same problem but in the case that instead od the partial
metric p, the standard metric d(x,y) = |x − y| is used (note that this is the so-
called associated metric of p, see, e.g., [1]). Take x = 1 and y = 1

2
. Then

ψ(d(Tx,Ty)) =

∣∣∣∣12 − 16
∣∣∣∣ = 13,

M(x,y) = max

{
1

2
,

1

2
,

1

3
,

1

2

(
5

6
+ 0

)}
=

1

2
,

α(M(x,y)) −β(M(x,y)) =
1

2
−

1
2

1 + 1
2

=
1

6
<

1

3
,

and the contractive condition (3.2) with p = d is not satisfied (neither is condition
(1.2) of Theorem 1.2).

Acknowledgements

The authors are thankful to the Ministry of Education, Science and Technological
Development of Serbia.

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1Faculty of Mathematics, University of Belgrade, Studentski trg 16, 11000 Beograd,

Serbia

2Faculty of Mechanical Engineering, University of Belgrade, Kraljice Marije 16,

11120 Beograd, Serbia

∗Corresponding author