EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 7124
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Fixed Point Theorems for (β, φ)-Expansive Mappings in
Controlled Metric Space

Manoj Kumar1, Neha Bhardwaj1, Gunaseelan Mani2, Rajagopalan Ramaswamy3,∗,
Khizar Hayat Khan3, Ola Ashour Abdelnaby3,4

1 Department of Mathematics, Maharishi Markandeshwar (Deemed to be University),
Mullana, Ambala 133207, Haryana, India
2 Department of Mathematics, Vel Tech Rengarajan Dr Sagunthala R & D Institute of
Sciences, Chennai 600062, Tamilnadu, India
3 Department of Mathematics, College of Science and Humanities,
Prince Sattam Bin Abdulaziz University, Alkharj 11942, Saudi Arabia
4 Department of Mathematics, Faculty of Science, University of Cairo, Egypt

Abstract. In the present manuscript, we have introduced a new notion of (β, φ)-expansive map-
pings in controlled metric space. In addition to this, some fixed point results are also proved with
the help of this notion. Some results from the literature are also deduced from our main results.
An example is also provided to prove the validity of our result. As an application an integral
equation is also solved with the help of our main result.

2020 Mathematics Subject Classifications: 47H10, 54H25

Key Words and Phrases: Controlled metric space, fixed point, (β, φ)-expansive mappings

1. Introduction

In 1922, Banach [1] proved an interesting result in fixed point theory known as Banach
contraction principle. This principle proved like a boom in the area of non-linear analysis
and its applications.

In 1989, Bakhtin [2] presented b-metric space, an extension of metric space and nu-
merous desirable fixed point solutions for expansive mappings were examined in b-metric
space. Also in 1993, Czerwik [3] expanded on the findings of b-metric space as well. The
idea of expansive mapping of Wang et al. [4] as follows “Let (X, d) be a metric space.
A mapping T : Ω → Ω on (Ω, d̄) such that ∀ u, e ∈ Ω : d̄(Tu, Te) ≥ d(u, e)”. After

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.7124

Email addresses: manojantil18@gmail.com (M. Kumar),
bhardwajneha15@gmail.com (N. Bhardwaj), mathsguna@yahoo.com (G. Mani),
r.gopalan@psau.edu.sa (R. Ramaswamy), k.khan@psau.edu.sa (K. H. Khan),
o.abdelnaby@psau.edu.sa (O. A. Abdelnaby)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 2 of 14

getting the motivation of this, Shahi et al. [5] proved fixed point theorems and gave some
applications for (ξ, α)-expansive mappings in complete metric spaces.

Recently, many research studies were conducted on b-metric space under different
expansive conditions.

After that, many authors used different types of contractive conditions including (α, ψ)-
expansive mappings in different metric and metric like spaces (see [6–14]).

In 2017, Kamran et al. [8] introduced an extended generalization of the b-metric space
and many results in the literature were generalized by their study. Mlaiki et al. [15] pro-
duced a controlled metric type space, which is an extension of the extended b-metric space.

2. Preliminaries

Definition 1 ([8]). Let Ω be a non-empty set and define the mappings µ : Ω×Ω → [1,∞)
and d̄ : Ω× Ω → [0,∞) such that for all u, e, f ∈ Ω

[label=(i),itemsep=-.16em,topsep=2pt]

(i) d̄(u, e) = 0 ⇐⇒ u = e,

(ii) d̄(u, e) = d̄(e, u),

(iii) d̄(u, e) ≤ µ(u, e)[d̄(u, f) + d̄(f, e)].

Then the pair (Ω, d̄) is known as an extended b-metric space. Further, Mlaiki et al. [15]
introduced the notion of controlled metric type space which is defined as follows:

Definition 2 ([15]). On a non-empty set Ω, define the mappings µ : Ω×Ω → [1,∞) and
d̄ : Ω× Ω → [0,∞) such that for all u, e, f ∈ Ω the following condition holds

[label=(s2),itemsep=-.16em,topsep=2pt]

(i) d̄(u, e) = 0 ⇐⇒ u = e,

(ii) d̄(u, e) = d̄(e, u),

(iii) d̄(u, e) ≤ µ(u, f)d̄(u, f) + µ(f, e)d̄(f, e).

Then the pair (Ω, d̄) is called a controlled metric type space.

To prove the above definition, we provide the following examples.

Example 1 ([15]). Choose Ω = {1, 2, 3, . . . }. Take d̄ : Ω× Ω → [0,∞) such that

d̄(u, e) =



0 ⇔ u = e

1

u
, if u = 2j and e = 2j + 1,

1

e
, if u = 2j + 1 and e = 2j,

1, otherwise.

(1)



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 3 of 14

Consider µ : Ω× Ω → [1,∞) as

µ(u, e) =


u, if u = 2j and e = 2j + 1,

e, if u = 2j + 1 and e = 2j,

1, otherwise.

(2)

It is clear that the conditions (s1) and (s2) are satisfied.
Now we investigate (s3).

[label=Case I:,itemsep=-.16em,topsep=2pt,itemindent=-1.75em,align=left]

(i) If f = u or f = e, (s3) is satisfied.

(ii) If f ̸= u and f ̸= e. (s3) holds when u = e.

Now we may assume that u ̸= e. Then, we have u ̸= e ̸= f . It is clear that (s3) holds in
each of the following subcases:

[label=(1),itemsep=-.16em,topsep=2pt]

(i) u, f are even and e = 2j + 1.

(ii) u = 2j are and e, f are odd.

(iii) u, f are odd and e = 2j.

(iv) u, e, f are even.

(v) u, e are even and f = 2j + 1.

(vi) u, e are odd and f = 2j.

(vii) u, f are odd.

Thus d̄ is a controlled metric type space.
Moreover for j = 2, 3, . . . we have

d̄(2j + 1, 4j + 1) = 1 >
1

j
µ(2j + 1, 4j + 1)[d̄(2j + 1, 2j) + d̄(2j, 4j + 1)]. (3)

Therefore d̄ is not an extended b-metric space.

Example 2 ([15]). Take Ω = {0, 1, 2}. Consider the function d̄ given as

d̄(0, 0) = d̄(1, 1) = d̄(2, 2) = 0,

d̄(0, 1) = d̄(1, 0),

d̄(0, 2) = d̄(2, 0) =
1

2
,

d̄(1, 2) = d̄(2, 1) =
2

5
. (4)



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 4 of 14

Define a symmetric function µ : Ω× Ω → [1,∞) such that

µ(0, 0) = µ(1, 1) = µ(2, 2) = µ(0, 2) = 1,

µ(1, 2) =
5

4
,

µ(0, 1) =
11

10
. (5)

One can easily verify that d̄ is a controlled metric type space.
Since

d̄(0, 1) = 1 >
99

100
µ(0, 1)[d̄(0, 2) + d̄(2, 1)] . (6)

d̄ is not an extended b-metric.

The concept of Cauchy and convergent sequences in controlled metric type spaces are
defined as follows:

Definition 3 ([15]). Let (Ω, d̄) be a controlled metric type space and {uj}j≥0 be a sequence
in Ω.

[label=(iii),itemsep=-.16em,topsep=2pt]

(i) The sequence {uj} converges to some u ∈ Ω if for all ε > 0 ∃ P = P (ε) ∈ N such
that d̄(uj , u) < ε for all j ≥ P , we write lim

j→0
uj = u.

(ii) We say that {uj} is Cauchy if for all ε > 0 ∃ P = P (ε) ∈ N such that d̄(ui, uj) < ε
for all j ≥ P .

(iii) If every Cauchy sequence is convergent then the space (Ω, d̄) is called complete.

Definition 4 ([15]). Let (Ω, d̄) be a controlled metric type space. Let u ∈ Ω and ε > 0,
C(u, ε) is defined as

[label=(iv),itemsep=-.16em,topsep=2pt]

(i) The open ball

C(u, ε) = {e ∈ Ω, d̄(u, e) < ε}. (7)

(ii) A self mapping T on Ω is said to be continuous at u ∈ Ω if ∀ ε > 0 ∃ δ > 0 such
that T (C(u, δ)) ⊆ C(Tu, ε).

Remark 1. If for all u, e in Ω, µ(u, e) = t ≥ 1, then it is a b-metric space. Therefore, we
conclude that every b-metric space is controlled metric space. However the converse is not
always true.

Clearly, If a mapping T is continuous at u in the controlled metric type space then
uj → u⇒ Tuj → u as j→ ∞



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 5 of 14

Definition 5 ([16]). Let Φ denote the set of all functions, φ : [0,∞) → [0,∞) such that

[label=(v),itemsep=-.16em,topsep=2pt]

(i) φ is non-decreasing.

(ii) For all q > 0 where φj is the jth iterate of φ.

Now, we recall the following lemma.

Lemma 1 ([5]). If φ ∈ Φ then φ(q) < q for all q ∈ (0,∞).

Next, Mlaiki et al. [? ] introduced the following notion.

Definition 6 ([15]). Let Ω be a non-empty set and µ : Ω × Ω → [1,∞) be a mapping. A
function φ : [0,∞) → [0,∞) is said to be controlled comparison function if φ satisfies the
following conditions

[label=(vi),itemsep=-.16em,topsep=2pt]

(i) φ is non-decreasing.

(ii)
∞∑
j=1

φj(q)
j∏

m=1
µ(um, ui)µ(uj , uj+1) < ∞ and lim

j→0
φj(q)µ(uj , uj+1) < ∞ for any se-

quence {uj} in Ω for all ε > 0 and non-negative integer i, φi is the ith iterate of
φ.

The set of all controlled comparison functions is denoted by Φ which is an extension of
b-comparison function of Berinde.

Note that if φ ∈ Φ then we have
∑∞

j=1 φ
j < ∞, since

∞∑
j=1

φj
j∏

m=1
µ(um, ui) ≥ φj(q) ∀

q ≥ 0. Hence by Lemma 1, we have φ(q) < q.

To show that the family Φ is a non-empty set. We present the following example.

Example 3 ([15]). Consider the controlled b-metric space (Ω, d̄) which was defined in
Example 2.

Define the mapping [17] φ : [0,∞) → [0,∞) by φ(q) =
( rq

2

)
where r < 1.

Note that µ(u, e) ≤ 2. Then, we have

φj(q)

j∏
m=1

µ(um, ui)µ(uj , uj+1) ≤
(
rjq

2j

)
· 2j+1 = 2rjq.

Therefore,
∞∑
j=1

φj(q)

j∏
m=1

µ(um, ui)µ(uj , uj+1) ≤
∞∑
j=1

2rjq <∞ .

Similarly, it is not difficult to see that lim
j→0

φj(q)µ(uj , uj+1) <∞.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 6 of 14

3. Main Result

In this section, we shall introduce a new notion of (β, φ)-expansive mapping in con-
trolled metric space and proved some fixed point result by making use of this notion.

Definition 7 ([16]). Let Φ denote all functions, φ : [0,∞) → [0,∞) which satisfy the
following properties

[label=(vii),itemsep=-.16em,topsep=2pt]

(i) φ is non-decreasing.

(ii)
∞∑
j=1

φj(a) <∞ for each a > 0 where φj is the jth iterate of φ.

Lemma 2 ([16]). If φ : [0,∞) → [0,∞) is a non-decreasing function then for each a > 0,

lim
j→0

φj(a) = 0

⇒ φ(a) < a

Definition 8 ([5]). Let (Ω, d̄) be a metric space and T : Ω → Ω be a given mapping. We
say that T is an (β, φ)-expansive mapping if ∃ two functions φ ∈ Φ and β : Ω×Ω → [0,∞)
such that

φ(d̄(Tu, Te)) ≥ β(u, e)d̄(u, e), ∀ u, e ∈ Ω (8)

Definition 9 ([5]). Let T : Ω → Ω and β : Ω × Ω → [0,∞). We say T is said to be β
admissible if ∀ u, e ∈ Ω with β(u, e) ≥ 1 then β(Tu, Te) ≥ 1.

Now we prove our first result.

Theorem 1. Let (Ω, d̄) be a complete controlled metric type space and T : Ω → Ω be a
bijective and (β, φ)-expansive mapping satisfying the following conditions:

[label=(i),itemsep=-.16em,topsep=2pt]

(i) T−1 is β admissible,

(ii) ∃ u0 ∈ Ω such that β(u0, T
−1u0) ≥ 1,

(iii) T is continuous.

Then T has a fixed point, i.e. ∃ u ∈ Ω such that Tu = u.
Moreover, if for any two fixed points of T in Ω say a, b, we have β(a, b) ≥ 1 then T

has a unique fixed point in Ω.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 7 of 14

Proof. Let u0 be the point in Ω satisfying condition (ii) in our theorem.
Define a sequence {uj} in Ω by Tuj = uj−1 ∀ j ∈ N .
First of all, note if ∃ j such that uj = uj+1, then we are done and uj is the fixed point

of T .
So we may assume that uj ̸= uj+1 ∀ j ≥ 0.
Also from the hypothesis of our theorem, we know that β(u0, T

−1u0) = β(u0, u1) ≥ 1,
using the result T−1 is β admissible. We can easily deduce that for all j ≥ 0 β(uj , uj+1) ≥ 1.

Now using the fact that T is (β, φ)-expansive mapping.
We deduce that

d̄(uj , uj+1) ≤ β(uj , uj+1)d̄(uj , uj+1)

≤ φ(d̄(Tuj , Tuj+1))

= φ(d̄(uj−1, Tuj)).

By repeating the same process, we get

d̄(uj , uj+1) ≤ φjt(d̄(u0, u1)), ∀ j ∈ N (9)

Hence for all i, j ∈ N with i > j, we have

d̄(uj , ui) ≤ µ(uj , uj+1)d̄(uj , uj+1) + µ(uj+1, ui)d̄(uj+1, ui)

≤ µ(uj , uj+1)d̄(uj , uj+1) + µ(uj+1, ui)µ(uj+1, uj+2)d̄(uj+1, uj+2)

+ µ(uj+1, ui)µ(uj+2, ui)d̄(uj+2, ui)

≤ µ(uj , uj+1)d̄(uj , uj+1) + µ(uj+1, ui)µ(uj+1, uj+2)d̄(uj+1, uj+2)

+ µ(uj+1, ui)µ(uj+2, ui)µ(uj+2, uj+3)d̄(uj+2, uj+3)

+ µ(uj+1, ui)µ(uj+2, ui)µ(uj+3, ui)d̄(uj+3, ui)

≤ µ(uj , uj+1)φ
j d̄(u0, u1) + µ(uj+1, uj+2)µ(uj+1, ui)φ

j+1d̄(u0, u1)

+ µ(uj+1, ui)µ(uj+2, ui)µ(uj+2, uj+3)φ
j+2d̄(u0, u1) + . . .

+ µ(uj+1, ui)µ(uj+2, ui)µ(uj+3, ui) . . . µ(ui−2, ui−1)φ
i−1d̄(u0, u1)

= µ(uj , uj+1)φ
j d̄(u0, u1) +

i−2∑
n=j+1

φnd̄(u0, u1)
n∏

m=j+1

µ(um, ui)µ(un, un+1)

= µ(uj , uj+1)φ
j d̄(u0, u1) +

i−2∑
n=1

φnd̄(u0, u1)
n∏

m=1

µ(um, ui)µ(un, un+1)

−
j∑

n=1

φnd̄(u0, u1)

n∏
m=1

µ(um, ui)µ(un, un+1)

= Si−2 − Sj (10)

where

Sj =

j∑
n=1

φnd̄(u0, u1)

n∏
m=1

µ(um, ui)µ(un, un+1). (11)



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 8 of 14

Since φ ∈ Φ, we deduce that lim
i,j→∞

[Si−2, Sj ] = 0

And

lim
j→∞

[µj , µj+1]φ
j d̄(u0, u1) <∞ . (12)

Thus the sequence {uj}j≥0 is a Cauchy sequence in Ω and (Ω, d̄) being the complete
controlled metric type space. The {uj} converges to some u ∈ Ω.

Also note that

d̄(u, T−1) ≤ µ(u, uj+1)d̄(u, uj+1) + µ(uj+1,T
−1u)d̄(uj+1, T

−1u)

= µ(u, u)d̄(u, u) + µ(u, T−1u)d̄(u, T−1u) . (13)

Proceeding the limit as j → ∞, we get d̄(u, uj+1) = 0.
And being T is continuous, we conclude that d̄(u, T−1u) = 0, i.e.,

T−1u = u

⇒ T (T−1u) = Tu

⇒ (TT−1)u = Tu

⇒ u = Tu.

Thus u is a fixed point T .
Now suppose that T has two fixed points a, b such that β(a, b) ≥ 1.
Now using the fact that T is an (β, φ)-expansive mapping and β-admisible, we obtain,

d̄(a, b) = d̄(T−1a, T−1b)

≤ β(a, b)d̄(T−1a, T−1b)

≤ φd̄(a, b)

...

≤ φj d̄(a, b) . (14)

Since φ ∈ Φ, taking the limit in the above inequality, we deduce that d̄(a, b) = 0 which
implies that a = b.

Thus T has a unique fixed point.
In the next theorem, we shall replace the continuity of T by the following weaker

condition.
If {uj}∞j=1 is a sequence in Z such that β(uj , uj+1) ≥ 1 for all j and uj → u as j → ∞

then

β(uj , u) ≥ 1 ∀ j.

Now we present an example to prove the validity of our result.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 9 of 14

Example 4. Let (Ω, d̄) be the controlled metric type space on Ω = {0, 1, 2}.
Define the function β : Ω× Ω → (−∞,∞) such that

β(u, e) =

{
1, if (u, e) = (1, 1),
1
20 , if (u, e) ̸= (1, 1).

(15)

Define the self mapping T on Ω and Ω = {0, 1, 2} by T (0) = 2, T (1)1, T (2) = 0 and the
function φ(q) = 1

8q and d̄ : Ω× Ω → [0,∞) defined by

d̄(0, 0) = d̄(1, 1) = d̄(2, 2) = 0,

d̄(0, 1) = d̄(1, 0),

d̄(0, 2) = d̄(2, 0) =
1

2
,

d̄(1, 2) = d̄(2, 1) =
2

5
.

We want to verify that T satisfies the conditions of Theorem 1.
It is clear that T is continuous for u0 = 1.
We have β(1, T−1(1)) = β(1, 1) = 1 ≥ 1.
So T is β-admissible.
Now we verify that T is (β, φ)-expansive mapping.
Note that β(u, e) = β(e, u)

[label=(iv),itemsep=-.16em,topsep=2pt]

(i) β(u, u)d̄(Tu, Tu) = 0 ≤ φ(d̄(u, u)) = φ(0) = 0 ∀ u ∈ Ω,

(ii) β(1, 0)d̄(T1, T0) = 1
20(d̄(1, 2)) =

1
50 ≤ φd̄(1, 2) = φ

(
2
5

)
= 1

20 .

(iii) β(1, 2)d̄(T1, T2) = 1
20(d̄(1, 0)) =

1
20 ≤ φd̄(1, 0) = φ(1) = 1

8 .

(iv) β(2, 0)d̄(T2, T0) = 1
20(d̄(0, 2)) =

1
40 ≤ φd̄(0, 2) = φ

(
1
2

)
= 1

16 .

Therefore, T satisfies the conditions in Theorem 1 and hence it has a unique fixed point
u = 1.

Theorem 2. Let (Ω, d̄) be a complete Controlled metric type space and be a T : Ω → Ω
be a bijective and (β, φ)-expansive mapping for some φ ∈ Φ. Suppose that the following
condition holds:

[label=(ii),itemsep=-.16em,topsep=2pt]

(i) T−1 is β-admissible,

(ii) ∃ u0 ∈ Ω such that β(u0, T
−1u0) ≥ 1,

(iii) If {uj}∞j=1 is a sequence in Ω such that β(uj , uj+1) ≥ 1 for all j and {uj} → u as

j → ∞ then β(T−1uj , T
−1u) ≥ 1 ∀ j.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 10 of 14

Then T has a fixed point.

Proof. In proving the result, we follow the same steps as in the proof of Theorem 1 to
construct a sequence {uj}∞j=1 that converges to a point u, u ∈ Ω. The constructed sequence
has the property β(uj , uj+1) ≥ 1, for all natural numbers j. The last assumption of the
result implies that β(T−1uj , T

−1u) ≥ 1.
Now, we will prove that u is a fixed point of T .
The triangle inequality implies that

d̄(T−1u, u) ≤ µ(T−1u, uj)d̄(T
−1u, uj) + µ(uj , u)d̄(uj , u)

= µ(T−1u, uj)d̄(T
−1u, uj) + µ(u, u)d̄(u, u)

= µ(T−1u, uj)d̄(T
−1u, uj) + 0.

Thus, we get

d̄(T−1u, u) ≤ µ(T−1u, uj)d̄(T
−1u, uj)

≤ µ(T−1u, uj)d̄(T
−1u, uj)β(T

−1u, uj)

≤ µ(T−1u, uj)φ(d̄(u, Tuj))

≤ µ(T−1u, u)φ(d̄(u, Tu)). (16)

Continuity of φ at u = 0 implies that d̄(T−1u, u) = 0 as j → +∞ i.e. T−1u = u.
Consider Tu = T (T−1u) = (TT−1)u = u.
Hence the result is proved.

Example 5. Let Ω = [0,∞) with controlled metric space d̄ : Ω × Ω → [0,∞) defined by
d̄ = |u − e| ∀ u, e ∈ Ω along with the mapping µ : Ω × Ω → [1,∞) and define T : Ω → Ω
is (β, φ)-expansive mapping with φ ∈ Φ by

T (u) =

{
u3, u ≥ 1

2 ,
u
2 , 0 ≤ u < 1

2

and

β(u, e) =

{
0, u, e ≥

[
0, 12

)
,

1, otherwise.

Clearly T is not continuous at 1
2 and T is (β, φ)-expansive mapping with φ(r) = r

2 ∀ r ≥ 0.
Moreover for all u, e ∈ Ω, we have

1

2
(d̄(Tu, Te)) ≥ β(u, e)d̄(u, e).

Also there exist u0 ∈ Ω such that β(u0, T
−1u0) ≥ 1. Infact for u0 = 1, we have β(1, T−11) =

1.
Now, let u, e ∈ Ω such that β(u, e) ≥ 1. This implies that u ≥ 1, e ≥ 1.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 11 of 14

And by definition of T−1 and β, we have

T−1u = u
1
3 ≥ 1, T−1e = e

1
3 ≥ 1 and β(T−1u, T−1) = 1

i.e. T−1 is β-admissible.
Finally, let {uj} be a sequence in Ω such that β(uj , uj+1) ≥ 1 ∀ j and {uj} → u ∈ Ω

as j → ∞.
Since β(uj , uj+1) ≥ 1 ∀ j by the definition of β, we have uj ≥ 1 ∀ j and u ≥ 1 then

β(T−1 uj , T
−1u) = 1.

Therefore, all the conditions of Theorem 2 are satisfied. So that T has the fixed points.
Here 0, 1 are two fixed points of T .

To ensure the uniqueness of the fixed point in Theorem 2, we consider the following
condition:

(N): For all g, e ∈ Ω, there exist w ∈ Ω such that β(g, w) ≥ 1 and β(e, w) ≥ 1.

Theorem 3. Adding the hypothesis of Theorem 2 to the condition (N), we yield the
uniqueness of the fixed point of T .

Proof. In this theorem, we will the prove of uniqueness of fixed points, i.e.

T (g) = g, T (e) = e, ∀ g, e ∈ Ω.

From the condition (N) there exist w ∈ Ω such that β(g, w) ≥ 1 and β(e, w) ≥ 1.
Using the β-admissible property of T−1, we get

β(g, T−1w) ≥ 1 and β(e, T−1w) ≥ 1.

Therefore by repeatedly using β-admissible property of T−1, we get

β(g, T−jw) ≥ 1 and β(e, T−jw) ≥ 1 ∀ j ∈ N. (17)

Using the inequality of equations (8) and (17), we get

(d̄(g, T−jw)) ≤ β(g, T−jw)d̄(g, T−jw)

≤ φ(d̄(Tg, T−j+1w))

= φ(d̄(g, T−j+1w))

≤ φ(d̄(g, T−j+1w)).

By the repetition of above inequality, we get

(d̄(g, T−jw)) ≤ φ(d̄(g, w)) ∀ j ∈ N.

Thus we have,
T−jw → g as j → ∞.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 12 of 14

Similarly we may obtain,
T−jw → e as j → ∞.

From the uniqueness of the limit of T−jw, we get

g = e.

Hence the proof.

4. Consequences

Now we present the following as an immediate consequence of Theorem 1.

Corollary 1. Let (Ω, d̄) be a complete controlled metric type spaces and T : Ω → Ω be a
bijective and β : Ω× Ω → [0,∞) mappings satisfying the following conditions:

[label=(i),itemsep=-.16em,topsep=2pt]

(i) T is continuous.

(ii) ∃ φ ∈ Φ such that φ(d̄(Tu, Te)) ≥ β(u, e)d̄(u, e) ∀ u, e ∈ Ω.

Then T has a unique fixed point.

Proof. Define the function β : Ω× Ω → [0,∞) by β(u, e) = 1.
Note that T−1 is β-admissible. Moreover, T satisfies all the conditions of Theorem 1,

so T has a unique fixed point.

5. Application

Consider the non-linear Volterra Integral Equation

u(q) =

∫ q

0

u(l)

1 + |u(l)|
dl, q ∈ [0, 2].

Let Ω = C([0, 2],R) space of continuous real-valued functions on [0, 2] and

d̄(u, e) = sup |u(q), e(q)|, ∀ q ∈ [0, 2].

Define T : Ω → Ω by

(Tu)q =

∫ q

0

u(l)

1 + |u(l)|
dl, q ∈ [0, 2].

Define β : Ω× Ω → [0,∞) by β(u, e) = 1
4 and φ : [0,∞) → [0,∞) by φ(q) = q

4 .
Define a symmetric function µ : Ω× Ω → [1,∞) by µ(u, e) = 2 ∀ u, e ∈ [0, 2].
Clearly φ(q) is an increasing and continuous function of q and φ(0) = 0 and φ(q) < q

∀ q > 0.



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 13 of 14

Now

d̄(u, e) = sup |u(q)− e(q)| ∀ q ∈ [0, 2]

= sup

∣∣∣∣(∫ q

0

u(l)

1 + |u(l)|
dl

∫ q

0

e(l)

1 + |e(l)|
dl

)∣∣∣∣
= sup |(Tu)(q)(Te)(q)|
≤ |(Tu)(q)(Te)(q)|
= d̄(Tu, Te)

⇒ d̄(u, e) ≤ d̄(Tu, Te).

Now

φ(d̄(Tu, Te)) =
d̄(Tu, Te)

4

≥ d̄(u, e)

4
= β(u, e) d̄(u, e)

⇒ φ(d̄(Tu, Te)) ≥ β(u, e)d̄(u, e).

Thus T : Ω → Ω is (β, φ)-expansive mapping. Thus T has atleast one fixed point. Clearly
T (0) = 0 and hence 0 is the unique fixed point of T .

6. Conclusion

Fixed point results in the setting of controlled metric space using (β, φ)-expansive
mappings have been established in this manuscript. The derived results have been sup-
plemented with suitable non trivial example and the result is applied to find analytical
solution of integral equation. It will be open problem to extend the result in the generalised
forms of controlled metric and metric like spaces and find application to integro-differential
equations.

Acknowledgements

”The authors extend their appreciation to Prince Sattam bin Abdulaziz University for
funding this research work through the project number (PSAU/2025/01/33096)”.

Author Contributions

Conceptualisation M Kumar, G Mani, Ramaswamy R, Methodology M Kumar, G
Mani Sofaware Neha B, Ola A A, Supervision G Mani, Ramaswamy R, Writing Orignial
Draft M Kumar, G Mani, Khizar HK, Editing M Kumar, Neha B, G Mani, Rajagopalan
R, Ola AA, Khizar HK



M. Kumar et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7124 14 of 14

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