EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 4, 2017, 908-915
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

An unified theorem for mappings in orbitally complete
partial metric spaces

Valeriu Popa1, Alina-Mihaela Patriciu2,∗

1 “Vasile Alecsandri” University of Bacău, Romania
2 Department of Mathematics and Computer Sciences, Faculty of Sciences and Environment,
“Dunărea de Jos” University of Galaţi, Romania

Abstract. In this paper an unified theorem for mappings in orbitally complete partial metric
spaces is proved. This theorem generalizes and proves Theorems 8 and 9 [8], Theorem 3.2 [10] and
Theorem 2.6 [7].

2010 Mathematics Subject Classifications: 54H25, 47H10
Key Words and Phrases: Fixed point, Orbitally complete, Partial metric space, Implicit relation

1. Introduction

In 1974, Ćirić [4] has first introduced orbitally complete metric spaces and orbitally
continuous functions.

Let f be a self mapping of a metric space (X, d). If x0 ∈ X, every Cauchy sequence
of the orbit Ox0(f) = {x0, fx0, f2x0, ...} is convergent to a point y ∈ X, then X is said to
be f - orbitally complete in x0. If f is orbitally complete at each x ∈ X, then X is said to
be f - orbitally complete. Every complete metric space is f - orbitally complete for every
function f . An orbitally complete metric space may not be a complete metric space [[17],
Example 2].

Let f be a self mapping of a metric space (X, d). Then, the mapping f is said to
be orbitally continuous at the point x ∈ X if fyn converges to fz for any subsequence
yn ∈ Ox(f) which converges to the point z ∈ X. The function f is said to be orbitally
continuous if it is orbitally continuous at each x ∈ X. Any continuous self mappings
of a metric space is orbitally continuous. An orbitally continuous mapping may not be
continuous [[17], Examples 4, 5].

Some fixed point results for mappings in orbitally complete metric spaces are obtained
in [2], [5], [11], [12] and in other papers.

In 1994, Matthews [9] introduced the concept of partial metric spaces as a part of the
study of denotional semantics of data flow net work and proved the Banach contraction

∗Corresponding author.

Email addresses: vpopa@ub.ro (V. Popa), Alina.Patriciu@ugal.ro (A.-M. Patriciu)

http://www.ejpam.com 908 c© 2017 EJPAM All rights reserved.



V. Popa, A.-M. Patriciu / Eur. J. Pure Appl. Math, 10 (4) (2017), 908-915 909

principle in such spaces. Many authors studied the fixed points for mappings satisfying
contractive conditions in complete partial metric spaces in [1], [3], [6] and in other papers.

Recently, in [8] the authors initiated the study of fixed points in orbitally complete
partial metric spaces.

In [7] and [10] new results are obtained.
Several classical fixed point theorems and common fixed point theorems have been

unified considering a general condition by an implicit relation in [13], [14] and in other
papers. Recently, the method is used in the study of fixed points in metric spaces, sym-
metric spaces, quasi - metric spaces, b - metric spaces, ultra - metric spaces, convex metric
spaces, reflexive spaces, compact metric spaces, paracompact metric spaces, in two and
three metric spaces, for single - valued mappings, hybrid pairs of mappings and set - valued
mappings.

Quite recently, the method is used in the study of fixed points for mappings satisfying
a contractive condition of integral type, in fuzzy metric spaces, probabilistic metric spaces,
intuitionistic metric spaces and G - metric spaces. With this method the proof of some
fixed point theorems is more simple. Also, the method allows the study of local and global
properties of fixed point structures.

The study of fixed points for mappings satisfying an implicit relation in orbitally metric
spaces is initiated in [15], [16] and in other papers. The study of fixed points for mappings
satisfying an implicit relation in partial metric spaces is initiated in [18].

The purpose of this paper is to prove a general fixed point theorem for self mappings
in orbitally complete partial metric spaces which generalizes and improves Theorem 2.6
[7], Theorem 8 and 9 [8] and Theorem 3.2 [10].

2. Preliminaries

Definition 1 ([9]). Let X be a nonempty set. A function p : X ×X → R+ is said to be
a partial metric on X if for any x, y, z ∈ X, the following conditions hold:

(P1) : p(x, x) = p(y, y) = p(x, y) if and only if x = y,
(P2) : p(x, x) ≤ p(x, y),
(P3) : p(x, y) = p(y, x),
(P4) : p(x, z) ≤ p(x, y) + p(y, z)− p(y, y).
The pair (X, p) is called a partial metric space. If p(x, y) = 0, then (P1) and (P2)

implies x = y, but the converse does not always hold.

Each partial metric space on X generates a T0 topology τp which has as base the
family of open p - balls {Bp(x, ε) : x ∈ X and ε > 0}, where Bp(x, ε) = {y ∈ X : p(x, y) ≤
p(x, x) + ε} for all x ∈ X and ε > 0.

If p is a partial metric on X, then the function dp(x, y) = 2p(x, y) − p(x, x) − p(y, y)
defines a metric on X. Further, a sequence (xn) converges in (X, dp) to a point x ∈ X if

lim
n,m→∞

p(xn, xm) = lim
n→∞

p(xn, x) = p(x, x).



V. Popa, A.-M. Patriciu / Eur. J. Pure Appl. Math, 10 (4) (2017), 908-915 910

Lemma 1 ([2], [9]). Let (X, p) be a partial metric space and xn a sequence in X convergent
to z, where p(z, z) = 0. Then, limn→∞ p(xn, y) = p(z, y) for every y ∈ X.

Definition 2 ([9]). Let (X, p) be a partial metric space.
a) A sequence {xn} in X is a Cauchy sequence if and only if limn,m→∞ p(xn, xm)

exists and is finite.
b) A partial metric space is said to be complete if every Cauchy sequence converges

with respect to τp to a point x ∈ X such that p(x, x) = limn,m→∞ p(xn, xm).

Remark 1. The following examples [9] show that a convergent sequence in partial metric
space may not be Cauchy. In particular, it is shown that the limit of a convergent sequence
is not unique.

Example 1 ([9]). Let p : R+ × R+ → R+ be a partial metric defined as p(x, y) =
max{x, y}. Define a sequence {xn} in X as

xn =

{
0, n = 2k
1, n = 2k + 1, k ∈ N.

Then {xn} is a convergent sequence but limn,m→∞ p(xn, xm) does not exists.

Definition 3 ([8]). Let (X, p) be a partial metric space. A mapping T : X → X is called
orbitally continuous if limi→∞ p(T

nix, z) = p(z, z) implies limi→∞ p(TT
nix, z) = p(Tz, z)

for each x ∈ X.

Definition 4 ([8]). A partial metric space is called orbitally complete if every Cauchy
sequence {Tnix}∞i=1 converges in (X, p), that is

lim
i,j→∞

p(Tnix, Tnjx) = lim
i→∞

p(Tnix, z) = p(z, z).

Theorem 2 ([8]). Let T : X → X be an orbitally continuous function on an orbitally
complete partial metric space (X, p). If

min{p (Tx, Ty) , p (x, Tx) , p (y, Ty)} ≤ ap (x, y)

for some 0 ≤ a < 1 and all x, y ∈ X, then the sequence {Tnx} converges to a fixed point
of T in X.

Theorem 3 ([8]). Let T : X → X be an orbitally continuous function on an orbitally
complete partial metric space (X, p). If

min{p (Tx, Ty) · p(x, y), p (x, Tx) · p (y, Ty)}
min{p (x, Tx) , p (y, Ty)}

≤ ap (x, y)

for some 0 ≤ a < 1 and all x, y ∈ X such that p(x, Tx) 6= 0 and p(y, Ty) 6= 0, then the
sequence {Tnx} converges to a fixed point of T .



V. Popa, A.-M. Patriciu / Eur. J. Pure Appl. Math, 10 (4) (2017), 908-915 911

Theorem 4 ([10]). Let (X, p) be an orbitally complete partial metric space and let T :
X → X be an orbitally continuous function that satisfy

p (Tx, Ty) ≤ ap (x, y) + b
p (x, Tx) + p (y, Ty)}

1 + p (x, y)

for all x 6= y, where a, b ≥ 0 and a+ b < 1. Then T has a fixed point z in X. Moreover,
p(z, Tz) = p(Tz, Tz) = p(z, z) = 0.

Theorem 5 ([7]). Let T : X → X be an orbitally continuous function on an orbitally
complete partial metric space. Suppose that

min{p2 (x, Tx) , p2 (y, Ty) , p (x, y) · p (Tx, Ty)} ≤ ap (x, Tx) · p (y, Ty)

for all all x, y ∈ X and for some 0 ≤ a < 1. Then for each x ∈ X, the sequence {Tnx}
converges to a fixed point of T .

3. Implicit relations

Definition 5. Let Fop be the sets of all continuous functions F (t1, t2, ..., t5) : R5
+ → R

satisfying the following conditions:
(F1) : F is not increasing in variable t5,
(F2) : There exists h ∈ (0, 1) such that for all u ≥ 0, v > 0, F (u, v, v, u, u + v) ≤ 0

implies u ≤ hv.

In the following examples, the condition (F1) is obviously.

Example 2. F (t1, ..., t5) = t1 − at2 − bt3 − ct4 − dt5, where a > 0, b, c, d ≥ 0 and
a+ b+ c+ 2d < 1.

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u − av − bv − cu − d (u+ v) ≤ 0.
Then u ≤ hv, where 0 < h = a+b+d

1−(c+d) < 1.

Example 3. F (t1, ..., t5) = t1 − kmax{t2, t3, t4, t5}, where k ∈
(
0, 12
)
.

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u − k (u+ v) ≤ 0. Then u ≤ hv,
where 0 < h = k

1−k < 1.

Example 4. F (t1, ..., t5) = t1 − kmax
{
t2, t3, t4,

t5
2

}
, where k ∈ (0, 1).

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u − kmax
{
u, v, u+v

2

}
≤ 0. If

u > v, then u(1 − k) ≤ 0, a contradiction. Hence, u ≤ v which implies u ≤ hv, where
0 < h = k < 1.

Example 5. F (t1, ..., t5) = t21−at2t3− bt24− ct25, where a > 0, b, c ≥ 0 and a+ b+ 4c < 1.
(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u2 − av2 − bu2 − c (u+ v)2 ≤ 0. If

u > v, then u2 [1− (a+ b+ 4c)] ≤ 0, a contradiction. Hence, u ≤ v which implies u ≤ hv,
where 0 < h =

√
a+ b+ 4c < 1.



V. Popa, A.-M. Patriciu / Eur. J. Pure Appl. Math, 10 (4) (2017), 908-915 912

Example 6. F (t1, ..., t5) = t21 + t1
1+t5

−
(
at22 + bt23 + ct24

)
, where a > 0, b, c ≥ 0 and

a+ b+ c < 1.
(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u2 + u

1+u+v −
(
av2 + bv2 + cu2

)
≤ 0,

which implies u2 −
(
av2 + bv2 + cu2

)
≤ 0. Hence, u ≤ hv, where 0 < h =

√
a+b
1−c < 1.

Example 7. F (t1, ..., t5) = t1−at2− b(1+t3)t4
1+t2

−ct5, where a > 0, b, c ≥ 0 and a+b+2c < 1.
(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = u − av − bu − c (u+ v) ≤ 0, which

implies u ≤ hv, where 0 < h = a+c
1−(b+c) < 1.

Example 8. F (t1, ..., t5) = min{t1, t3, t4} − at2 − bmin{t3, t5}, where a, b ≥ 0 and 0 <
a+ b < 1.

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = min{u, v} − av − bv ≤ 0. If u > v,
then v (1− (a+ b)) ≤ 0, a contradiction. Hence, u ≤ v which implies u ≤ hv, where
0 < h = a+ b < 1.

Example 9. F (t1, ..., t5) = min{t1t2, t3t4}−at2 min{t3, t4}− bmin{t22, t25}, where a, b ≥ 0
and 0 < a+ b < 1.

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = uv−avmin{u, v}−bmin{v2, (u+ v)2} ≤
0. If u > v, then v2 (1− (a+ b)) ≤ 0, a contradiction. Hence, u ≤ v which implies u ≤ hv,
where 0 < h =

√
a+ b < 1.

Example 10. F (t1, ..., t5) = min{t23, t1t2, t24}−at3t4−bt25, where a, b ≥ 0 and 0 < a+4b <
1.

(F2) : Let u ≥ 0, v > 0 and F (u, v, v, u, u+ v) = min{v2, uv, u2}−auv−b (u+ v)2 ≤ 0.
If u > v, then v2 (1− (a+ 4b)) ≤ 0, a contradiction. Hence, u ≤ v which implies u ≤ hv,
where 0 < h =

√
a+ 4b < 1.

Example 11. F (t1, ..., t5) = min{t1t3, t2t4}−at2 min{t3, t4}−bmin{t23, t25}, where a, b ≥ 0
and 0 < a+ b < 1.

The proof is similar to the proof of Example 9.

4. Main result

Theorem 6. Let (X, p) be an orbitally complete metric space and T : X → X be orbitally
continuous such that

F (p (Tx, Ty) , p (x, y) , p (x, Tx) , p (y, Ty) , p (x, Ty) + p (y, Tx)) ≤ 0 (1)

for all x 6= y ∈ X and F ∈ Fop. Then, T has an fixed point z such that p (z, z) =
p (Tz, Tz) = p (z, Tz) = 0.

Proof. Let x0 ∈ X and xn+1 = Tnx0 and so xn+1 = Txn. If there exists n ∈ N
such that xn = xn+1 = Txn, then xn is a fixed point. Suppose that xn 6= xn+1. Hence,
p (xn, xn+1) > 0 for all n ∈ N . By (1) we have successively

F (p (Txn, Txn+1) , p (xn, xn+1) , p (xn, Txn) ,
p (xn+1, Txn+1) , p (xn, Txn+1) + p (xn+1, Txn)) ≤ 0,



V. Popa, A.-M. Patriciu / Eur. J. Pure Appl. Math, 10 (4) (2017), 908-915 913

F (p (xn+1, xn+2) , p (xn, xn+1) , p (xn, xn+1) ,
p (xn+1, xn+2) , p (xn, xn+2) + p (xn+1, xn+1)) ≤ 0.

By (P3):

p (xn, xn+2) ≤ p (xn, xn+1) + p (xn+1, xn+2)− p (xn+1, xn+1) .

Then by (F1) we obtain

F (p (xn+1, xn+2) , p (xn, xn+1) , p (xn, xn+1) ,
p (xn+1, xn+2) , p (xn, xn+1) + p (xn+1, xn+2)) ≤ 0.

By (F2) we obtain
p (xn+1, xn+2) ≤ hp (xn+1, xn) .

For m > n, using (P4) we obtain

p (xn, xm) ≤ p (xn, xn+1) + ...+ p (xn+m−1, xn+m)

≤
(
hn + hn+1 + ...+ hm−1

)
p (x0, x1)

≤ hn

1− h
p (x0, x1) .

Thus, limn,m→∞ p (xn, xm) = 0. Hence, {xn} is a Cauchy sequence in (X, p) and
since (X, p) is orbitally complete, then {Tnx0} converges to a limit z ∈ X such that
limn→∞ p (Tnx0, T

mx0) = limn→∞ p (Tnx0, z) = p (z, z) = 0. Since T is orbitally continu-
ous, limn→∞ p (Tnx0, z) = p (z, z) implies

lim
n→∞

p
(
Tn+1x0, T z

)
= p (Tz, Tz) .

On the other hand

p (z, Tz) ≤ p
(
z, Tn+1x0

)
+ p

(
Tn+1x0, T z

)
= p (z, xn+2) + p

(
Tn+1x0, T z

)
.

Using Lemma 1 and letting n tends to infinity we obtain

p (z, Tz) ≤ p (Tz, Tz) .

By (1) we have

F (p
(
Tz, T 2z

)
, p (z, Tz) , p (z, Tz) ,

p
(
Tz, T 2z

)
, p
(
z, T 2z

)
+ p (Tz, Tz)) ≤ 0.

Then by (P4),

p
(
z, T 2z

)
≤ p (z, Tz) + p

(
Tz, T 2z

)
− p (Tz, Tz) .



REFERENCES 914

By (F1) we obtain

F (p
(
Tz, T 2z

)
, p (z, Tz) , p (z, Tz) ,

p
(
Tz, T 2z

)
, p (z, Tz) + p

(
Tz, T 2z

)
) ≤ 0

which implies by (F2) that
p
(
Tz, T 2z

)
≤ hp (z, Tz) .

Hence, by (P2) we obtain

p (z, Tz) ≤ p (Tz, Tz) ≤ p
(
Tz, T 2z

)
≤ hp (z, Tz)

which implies
p (z, Tz) (1− h) ≤ 0,

i.e.
p (z, Tz) = 0.

Hence, z = Tz and z is a fixed point of T .
Since

p (z, Tz) ≤ p (Tz, Tz) ≤ p (z, Tz) .

Hence
p (z, Tz) = p (Tz, Tz) .

Therefore
p (z, z) = p (z, Tz) = p (Tz, Tz) = 0.

Remark 2. a) By Theorem 6 and Example 8 with b = 0, we obtain a generalization
of Theorem 2.

b) By Theorem 6 and Example 9 we obtain Theorem 3.
c) By Theorem 6 and Example 7 with c = 0, we obtain Theorem 4.
d) By Theorem 6 and Example 10 we obtain Theorem 5.

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