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Fixed Point Theorem in Controlled Metric Spaces
Shyam Sunder Prasad Singh
Assistant Professor, Department of Mathematics S.N.Sinha College, Warisaliganj, Nawada, 805130,
India
ABSTRACT
In this paper, we present a new fixed point theorem that generalizes the result of Souayah and Hidri [5 ] within
the context of the generalized controlled metric space with three control functions.
Keywords: Caristi contraction, Controlled metric space, Double controlled metric space.
INTRODUCTION
Fixed point theory has numerous applications in science and engineering. The proof of existence and uniqueness
theorems for solutions of ordinary and boundary value problems often rely on the application of various fixed
point theorems. Among them, the Banach contraction principle is the most widely used, and it has been
extensively generalized by modifying the contractive condition or by extending the underlying metric
space.Kamran et al. [1 ] originate the theory of an extended b-metric space. Mlaiki et al. [ 2] originate the theory
of a controlled metric space. Abdeljawad et al. [3 ] further extended this idea by addressing the double controlled
metric space, while Tasneem et al. [ 4] advanced the theory by defining a triple controlled metric space. Recently,
Souayah and Hidri [5 ] established a new fixed point theorem inspired by Caristi’s contraction principle within
the scheme of controlled metric spaces. Furthermore, Singh et al. [6 ] originate a generalization of the controlled
metric space by integrating three control functions. In this paper, we present a new fixed point theorem that
generalizes the result of Souayah and Hidri [5 ] within the context of the generalized controlled metric space
with three control functions.
Preliminaries
Definition 2.1 [1] Let Ɀ ≠ Ф, d : Ɀ х Ɀ → [0,∞) be a non – negative function and α : Ɀ х Ɀ → [1,∞) be control
functions satisfying the following conditions:
I. d(f,g) = 0 if and only if f = g,
II. d(f,g) = d(g,f)
III. d(f,g) ≤ α(f,g) [d(f,h) + d(h,g) ] ꓯ f,g,h ε Ɀ. Then
d is called a extended b- metric and (Ɀ, d) is a extended b-metric space.
Definition 2.2 [2] Let Ɀ ≠ Ф, d : Ɀ х Ɀ → [0,∞) be a non – negative function and α : Ɀ х Ɀ → [1,∞) be control
functions satisfying the following conditions:
I. d(f,g) = 0 if and only if f = g,
II d(f,g) = d(g,f)
III. d(f,g) ≤ α(f,h) d(f,h) + α(h,g) d(h,g) ꓯ f,g,h ε Ɀ. Then
d is called a controlled metric and (Ɀ,d) is a controlled metric space.
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Example 2.3[2] Let Ɀ = [0,∞) and a function define by d : Ɀ х Ɀ → [0,∞) such that
d(f,g) = 0 if f = g,
=
1
𝑓
, if f is even and g is odd ,
=
1
𝑔
, if f is odd and g is even,
= 1 otherwise,
and the controlled function α : Ɀ х Ɀ → [1,∞)
α(f,g) = f, if f is even and g is odd
= g, if f is odd and g is odd
= 1, otherwise.
Then d is called a controlled metric and (Ɀ, d) is a controlled metric space.
Theorem 2.4 [5] Let (Ɀ,d,α) be complete controlled metric space and Ⱨ: Ɀ→Ɀ a function such
that for all f,g ε Ɀ such that
d(Ⱨg,Ⱨg,) ≤ ( £(f)- £(Ⱨf)) d(f, g) Where
£: Ɀ→R is bounded below function.
Suppose f
0
ε Ɀ, the sequence {f
n
} define by f
n
= Ⱨ
n
f
0
satisfies
lim
n→∞
α(f,f
n
,) lim
n→∞
α(f
n
,f) exists and are finite for every f ε Ɀ and the controlled function condition holds
sup
m≥1
lim
i→∞
α(f
i+1,
f
i+2
) α(f
i+1,
f
m
)/ α(f
i,
f
i+1
) < 1/k where k ε (0,1). Then
Ⱨ has a unique fixed point in Ɀ.
Example 2.5 [5] Let Ɀ= [0,1] , d(f,g) = │f-g│
2
and α(f,g) = f + g +2. It is easy to prove that (Ɀ,d,α) is complete
controlled metric space but not traditional metric space or b- metric space. Let
the function Ⱨ: Ɀ→Ɀ and £ : Ɀ→R defined by Ⱨ(f) = f/5+1/2 and £(f) = 2f
2
+ 3f.
(£(f)- £(Ⱨf))d(f,g) = 2f
2
+ 3f - { 2(f/5+1/2)
2
+ 3( f/5+1/2)}│f-g│
2
= (48f
2
/25 +2f-2) │f-g│
2
However, the original problem restates the whole expression as
(£(f)- £(Ⱨf))d(f, g) = (52f
2
/25 +4f+2) │f - g│
2
d(Ⱨf,Ⱨg) = │f - g│
2
/25. Hence,
d(Ⱨf,Ⱨg) ≤ (£(f)- £(Ⱨf))d(f, g)
Finally Ⱨ has a unique fixed point is Ⱨ( 5/8) = 5/8 ε [0,1].
Definition 2.6 [3] Let Ɀ ≠ Ф, d : Ɀ х Ɀ → [0,∞)be a non- negative function and α, β : Ɀ х Ɀ → [1,∞) be two
control functions satisfying the following conditions:
I. d(f,g) = 0 if and only if f = g,
II. d(f,g) = d(g,f)
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III. d(f,g) ≤ α(f,h)d(f,h) + β(h,g)d(h,g) ꓯ f,g,h ε Ɀ. Then
d is called a double controlled metric and (Ɀ,d) is a double controlled metric space.
Example2.7 [3]. Let Ɀ = [0,∞) and a function define d : Ɀ х Ɀ→ [0,∞) such that
d(f,g) = 0, if x = y,
=
1
𝑓
, if f ≥ 1 and g ε [0,1)
=
1
𝑔
, if g ≥ 1 and f ε [0,1)
= 1, otherwise
and the controlled functions α, β : Ɀ х Ɀ → [1,∞) by
α(f,g) = f, if f, g ≥ 1 and β(f,g) = 1, if f,g < 1
= 1, otherwise = max{f,g}, otherwise.
Then d is a double controlled metric and (Ɀ,d) is a double controlled metric space.
Singh et al. introduce a generalization of the controlled metric space by introducing three control functions.
Definition 2.8 [6] Let Ɀ ≠ Ф, d : Ɀ х Ɀ х Ɀ → [0,∞)be a non- negative function and α,β,γ : Ɀ х Ɀ х Ɀ → [1,∞)
be three control functions satisfying the following conditions:
I. d(f,g,h) = 0 if at least two of the three points are the same.
II. For f,g ε Ɀ such that f ≠ g there exist a point h ε Ɀ such that d(f,g,h) ≠ 0.
III. For f,g,h ε Ɀ, d(f,g,h) = d(f,h,g) = d(g,h,f) = d(h,f,g) = d(g,f,h) = d(h,g,f)
IV. For f,g,h,a ε Ɀ, d(f,g,h) ≤ α(f,g,a)d(f,g,a) + β(g,h,a)d(g,h,a) + γ(h,f,a)d(h,f,a).
Then d is called controlled metric and (Ɀ,d) is a controlled metric space.
Example 2.9 Let Ɀ = (0,1) and d : Ɀ х Ɀ х Ɀ → [0,∞)is defined by
d(f,g,h) = 0 , if at least two of the three points are the same.
= μ(f,g,h) e
1/2│f-g│+ 3/8│g-h│+ 1/8│h-f│
otherwise. For
continuous α(f,g,h) ≥ 0 symmetric in f,g,h. If suffices to only verify property
(IV) of definition 2.8.Which typically in such generalized setting takes a form like.
For all f,g,h,t ε Ɀ .
d(f,g,h) = α(f,g,t) d(f,g,t) + β(g,h,t) d(g,h,t) + γ(h,f,t) d(h,f,t)
for some control function α,β,γ : Ɀ х Ɀ х Ɀ → [1,∞).
Using Jensen
,
s inequality applied of the convex function f(x) = e
x
along with modulus power │ a│.
d(f,g,h) = μ(f,g,h) e
1/2│f-g│+ 3/8│g-h│+ 1/8│h-f│
≤
μ(f,g,h) [1/2 e
│f-g│
+
3/8 e
│g-h│
+1/8 e
│h-f│
]
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= μ(f,g,h) [1/2 e
1/2│f-g│+1/2│f-g│
+
3/8 e
1/2│g-h│+1/2│g-h│
+1/8 e
1/2│h-f│+1/2│h-f│
]
≤ [ 1/2e
1/2│f-g│
+1] μ(f,g,h) e
1/2│f-g│+3/8│g-t│+1/8│t-f│
+ [ 3/8e
1/2│h-g│
+1] μ(f,g,h) e
1/2│h-g│+3/8│g-t│+1/8│t-h│
+ [ 1/8e
1/2│h-f│
+1] μ(f,g,h) e
1/2│h-f│+3/8│f-t│+1/8│t-h│
=
α(f,g,t) d(f,g,t)
+ β(h,g,t) d(h,g,t) + γ(h f,t) d(h,f,t)
Where α(f,.g,t) = 1/2 e
1\2│f-g│
+ 1 ≥ 1
β(g,h,t) = 3/8 e
1\2│g-h│
+ 1 ≥ 1
γ(h,f,t) = 1/8 e
1\2│h-f│
+ 1 ≥ 1.
Definition 2.10 [ 6 ] Let (Ɀ,d) be a controlled metric space. Let Ⱨ : Ɀ →Ɀ we say Ⱨ is continuous at the pair
(a,b) ε Ɀ х Ɀ if for every ε > 0, there exist r > 0 such that
Ⱨ ( B
r
(f,g)) ϲ B
ϲ
(f,g).
Definition 2.11 [ 6 ] Let (Ɀ,d) be a controlled metric-type space.
I. The sequence {f
n
]
n ε N
converges to f ε Ɀ. If for every ε > 0, there exists an integer n ε N such that for all
n ≥ N. d(f
n
, f, t) < ε for some fixed t ε Ɀ.
II. The sequence {f
n
]
n ε N
is a Cauchy sequence,. If for every ε > 0, there exist an integer n ε N such that for all
n,m ≥ N d(f
n
, f
m
, t) < ε for some fixed t ε Ɀ.
III. The space (Ɀ,d) is complete, if every Cauchy sequence in Ɀ is convergent.
In order to simplify results, we define the following products as follows:
Define
U
0
( t, f
n
, f
n+1
) = α
U
1
(f
m
, t , f
n+1
, f
n+2
) = β γ
U
2
(f
m
, t , f
n+1
, f
n+2
, f
n+3
) = β α γ
U
3
(f
m
, t , f
n+1
, f
n+2
, f
n+3
, f
n+4
) = β α α
γ
In general
U
m-n-1
(f
m
, t , f
n+1
, f
n+2
, f
n+3
, f
n+4
…, f
m-1
) = β (α… (m-n-2)) times)
γ
Define
V
0
( t, f
n
, f
m
) = γ
V
1
(f
m
, t , f
n+1
, f
n+2
) = β β
V
2
(f
m
, t , f
n+1
, f
n+2
, f
n+3
) = β α β
V
3
(f
m
, t , f
n+1
, f
n+2
, f
n+3
, f
n+4
) = β α α
β
In general
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V
m-n-1
(f
m
, t , f
n+1
, f
n+2
, f
n+3
, f
n+4
…, f
m-1
) = β (α… (m-n-2)) times)
β
.
Main Result
Theorem 3.1 Let Ɀ ≠ Ф and (Ɀ,d) be a complete controlled metric space. Let
Ⱨ : Ɀ →Ɀ be a functions satisfying
(3.1) d(Ⱨf,Ⱨg,t) ≤ ( £(f)- £(Ⱨf))d(f,g.t) ꓯ f,g,t ε Ɀ.
Where £: Ɀ→R is bounded function.
Assume that
(3.2) sup
m≥1
lim
j→∞
I U
j+1
(f
m
, t , f
n+1
,… f
j+2
)/ U
j
(f
m
, t , f
n+1
,… f
j+1
) I <1/k
(3.3) sup
m≥1
lim
j→∞
I V
j+1
(f
m
, t , f
n+1
,… f
j+2
)/ V
j
(f
m
, t , f
n+1
,… f
j+1
) I <1/k, where kε (0,1)
Also assume that for each f ε Ɀ, the limits
(3.4) lim
n→∞
α(f,f
n
,t), lim
n→∞
β(f,f
n
,t) and lim
n→∞
γ(f,f
n
,t) exists.
Then Ⱨ has a unique fixed point in Ɀ.
Proof Let f
0
ε Ɀ be arbitrary. Define sequence {f
n
}
n ε N
by f
n+1
= Ⱨf
n
for n ε N. Then from (3.1), we get
d(f
n
,f
n+1
,t) = d(Ⱨf
n-1
,Ⱨf
n
,t) ≤ ( £(f
n-1
)- £(Ⱨf
n-1
)) d(f
n-1
,f
n
.t)
= ( £(f
n-1
)- £(f
n
)) d(f
n-1
,f
n
.t)
Let, b
n
= d(f
n-1
,f
n
.t).Then
(3.5) b
n+1
/ b
n
≤ £(f
n-1
)- £(f
n
) ꓯ n ε N.
This implies that the sequence { £(f
n
)} is non- increasing and bounded below hence convergent. Thus,
lim
n→∞
£(f
n
) = r > 0. Summing the inequality (3.5), we get
Ʃ
i=1
n
b
i+1
/b
i
≤ Ʃ
i=1
n
( £(f
i-1
)- £(f
i
))
= £(f
0
)- £(f
n
),
which is finite,so
(3.6) lim
i→∞
b
i+1
/b
i
= 0.
Hence there exists i
0
ε N such that for all i ≥ i
0
( 3.7) b
i+1
/b
i
≤ k for k ε (0,1), so
(3.8) d(f
i
,f
i+1
,t) ≤ k d(f
i-1
,f
i
,t) for all i ≥ i
0
By recursion
(3.9) d(f
i
,f
i+1
,a) ≤ k
i
d(f
0
,f
1
,t) for all i ≥ i
0
Showing Cauchy Property
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Let m,n εN with m> n , we obtain
d(f
n
,f
m
,t) ≤ α(f
n
,f
m
,f
n+1
) d(f
n
,f
m
,f
n+1
) + β(f
m
,t,f
n+2
) d(f
m
,t,f
n+2
) + γ(t, f
n
,f
n+1
) d(t, f
n
,f
n+1
)
≤ α(f
n
,f
m
,f
n+1
) d(f
n
,f
m
,f
n+1
) + β(f
m
,t,f
n+1
) α(f
n
,f
m
,f
n+1
) d(f
m
,t,f
n+1
)
+ β(f
m
,t,f
n+1
) β(t,f
n+1
,f
n+2
) d(t,f
n+1
,f
n+2
)+ β(f
m
,t,f
n+1
) γ(f
n+1
,f
m
,f
n+2
) d(f
n+1
,f
m
,f
n+2
)
+ γ(t, f
n
,f
n+1
) d(t, f
n
,f
n+1
)
≤ α(f
n
,f
m
,f
n+1
) d(f
n
,f
m
,f
n+1
) + β(f
m
,t,f
n+1
) α(f
m
,t,f
n+2
) α(f
m
,t ,f
n+3
) d(f
m
,t ,f
n+3
)
+β(f
m
,t,f
n+1
) α(f
m
,t,f
n+2
)β(t,f
n+2
,f
n+3
) d(t,f
n+2
,f
n+3
)
+β(f
m
,t,f
n+1
) α(f
m
,t,f
n+2
) γ(f
n+2
,f
m
,f
n+3
) d(f
n+2
,f
m
,f
n+3
)
+β(f
m
,t,f
n+1
)β(t,f
n+1
,f
n+2
)d(t,f
n+1
,f
n+2
)+ β(f
m
,t,f
n+1
) γ(f
n+1
,f
m
,f
n+2
) d(f
n+1
,f
m
,f
n+2
)
+ γ(t, f
n
,f
n+1
) d(t, f
n
,f
n+1
)
d(f
n
,f
m
,t) ≤ α(f
n
,f
m
,f
n+1
) d(f
n
,f
m
,f
n+1
) + β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) α (f
m
,t,f
n+4
) d(f
m
,t, f
n+4
)
+ β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) β (t,
f
n+3
,f
n+4
) d(t,f
n+3
, f
n+4
)
+ β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) γ(
f
n+3
,f
m
, f
n+4
) d(
f
n+3
,f
m
, f
n+4
)
+ β (f
m
,t,f
n+1
) β (f
m
,t,f
n+2
) β (t, f
n+2
, f
n+3
) d(
t,f
m
, f
n+3
) (
3.10) + β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) γ(
f
n+2
,f
m
, f
n+3
) d(
f
n+2
,f
m
, f
n+3
)
+ β (f
m
,t,f
n+1
) β (t, f
n+1
,f
n+2
) d(
t,f
n+1
, f
n+2
)
+ β(f
m
,t,f
n+1
) γ(f
n+1
, f
m
,f
n+2
) d(f
n+1
, f
m
,f
n+2
)
+ γ(t, f
n
,f
n+1
) d(t, f
n
,f
n+1
)
Using 3.9, we get
d(f
n
,f
m
,t) ≤ α (f
n
,f
m
,f
n+1
)k
n
d(f
0
,f
m
,f
1
)+β(f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) α (f
m
,t,f
n+4
) d(f
m
,t, f
n+4
)
+ β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) β (t,
f
n+3
,f
n+4
)k
n+3
d(t,f
0
, f
1
)
+ β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) α (f
m
,t,f
n+3
) γ(
f
n+3
,f
m
, f
n+4
)k
n+3
d(
f
0
,f
m
, f
1
)
+ β (f
m
,t,f
n+1
) α (f
m
,t,f
n+2
) β (t, f
n+2
, f
n+3
)k
n+2
d(
t,f
0
, f
1
)
(3,11) + β (f
m
,t,f
n+1
) α(f
m
,t,f
n+2
) γ(
f
n+2
,f
m
, f
n+3
)k
n+2
d(
f
0
,f
m
, f
1
)
+ β (f
m
,t,f
n+1
) β (t, f
n+1
,f
n+2
)k
n+1
d(
t,f
0
, f
1
)
+ β (f
m
,t,f
n+1
) γ(f
n+1
, f
m
,f
n+2
)k
n+1
d(f
0
, f
m
,f
1
)
+ γ(t, f
m
,f
n+1
)k
n
d(t, f
0
,f
1
).
It follows that inequality, (3.11), can be written as
d(f
n
,f
m
,t) ≤ k
n
d(f
0
,f
1
,f
m
) [ U
0
( f
n
,f
m,
f
n+1
) + k U
1
( f
m
,t,f
n+1,
f
n+2
) + k
2
U
2
( f
m
,t,f
n+1,
f
n+2
, f
n+3
) +…
+ k
m-n-1
U
m-n-1
( f
m
,t,f
n+1,
f
n+2
,.. f
m-1
…) ]
INTERNATIONAL JOURNAL OF RESEARCH AND SCIENTIFIC INNOVATION (IJRSI)
ISSN No. 2321-2705 | DOI: 10.51244/IJRSI |Volume XII Issue VIII August 2025
Page 165
www.rsisinternational.org
(3.12) + k
n
d(f
0
,f
1
,t) [V
0
( f
n
,f
m,
f
n+1
) + k V
1
( f
m
,t,f
n+1,
f
n+2
) + k
2
V
2
( f
m
,t,f
n+1,
f
n+2
, f
n+3
) +…
+ k
m-n-1
V
m-n-1
( f
m
,t,f
n+1,
f
n+2
,.. f
m-1
…)]
The ratio test together with 3.2 and3.3, implies that the series converges and thus the sequence {f
n
}
n εN
is a
Cauchy sequence. Since (Ɀ,d) is complete controlled metric space there exists f* ε Ɀ such that
d(f
n
,f*, t) →0 as n→∞.
Proving f* is a fixed point.Using (3.9) we get
, d(f*,Ⱨf*,t) ≤ α(f*,Ⱨf*,f
n+1
) d(f*,Ⱨf*,f
n+1
) + β(Ⱨf*,t,f
n+1
) d(Ⱨf*,t,Ⱨf*,f
n+1
) + γ(t,f*,f
n+1
) d(t,f*,f
n+1
)
≤ α (f*,Ⱨf*,f
n+1
) d(f*,Ⱨf*,f
n+1
) + β (Ⱨf*,t,f
n+1
) d(Ⱨf*,t,Ⱨf
n
) + γ(t,f*,f
n+1
) d(t,f*,f
n+1
)
≤ α (f*,Ⱨf*,f
n+1
) d(f*,Ⱨf*,f
n+1
) + β (Ⱨf*,t,f
n+1
) kd(f*,t,f
n
) + γ(t,f*,f
n+1
) d(t,f*,f
n+1
).
Taking the limit as n→∞and using (3.9), we get
d(f*,Hf*,t) ≤ 0.
Implies,Ⱨf* = f*.
Uniqueness
Suppose there is another fixed point f** ≠ f*. Then
(3.13) d(f*,f**,t) = d(Ⱨf*,Ⱨf**,t) ≤ (£(f*)- £(f*)) d(f*,f**,t)
d(f*,f**,t) = 0,
which implies f* = f**. Thus the mapping Ⱨ has a unique fixed point f* ε Ɀ.
Remark: Our result extend the corresponding result of Souayah and Hidri [5 ] and others.
REFERENCES
1. T. Kamran, M. samreen, Q.Ul Ain “ A Generalization of b- Metric space and Some Fixed Point
Theorems.” Mathematics 5(2017), 19. https://doi.org/ 10.3390/ math5020019.
2. N. Mlaiki, H. Aydi, N. Souayah, T. Abdeljawad “Controlled Metric Type Space and the Related
Contractions Principle. Mathematics,6(2018) 194. https://doi.org/10.3390/math 6100194.
3. T. Abdeljawad, N. Mlaiki, H. Aydi, N. Souayah “ Double Controlled Metric Type Spaces and Fixed
Point Results, Mathematics 6( 2018) 320. https://doi.org/ 10.3390/math 6120320.
4. Z,S. Tasneem, K. Gopalan, T.Abdeljawad “ A Different Approch to Fixed point Theorems on Triple
Controlled Metric Type Spaces with a Numerical Experiment, Dynamic System and Applications “
(2021)30, no.1, 111- 130. https:// doi.org/ 10.46719/dsa20213018.
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