Coincidence Point Results in Hausdorff Rectangular Metric Spaces with
an Application to Lebesgue Integral Function
1Department of Mathematics, Faculty of Science and Technology
Rambhai Barni Rajabhat University, Chanthaburi 22000, THAILAND
2Department of Mathematics and Computer Science
Faculty of Science and Technology
Rajamangala University of Technology Thanyaburi (RMUTT)
Thanyaburi, Pathumthani 12110, THAILAND
Abstract: -In this paper, we were able to produce certain coincidence point results for g-nondecreasing self-
mappings fulfilling certain rational type contractions in a Hausdorff rectangular metric space utilizing C-functions
and generalized (ϑ, φ)-contractive mappings obeying an admissibility-type assumption.
Key-Words: Rational contractions, Rectangular metric space, Lebesgue integral function
Received: September 21, 2021. Revised: May 17, 2022. Accepted: June 20, 2022. Published: July 14, 2022.
1 Introduction
The Banach contraction argument [1] is a significant
development in fixed point theory. In a variety of
ways, it has already been generalized and expanded.
In the literature, we encountered several innovative
types of metric spaces, such as the one proposed by
Branciari [2] and demonstrated an analogue of the Ba-
nach contraction principle in a rectangle metric space
by replacing the triangle inequality with a weaker hy-
pothesis termed quadrilateral inequality. Many au-
thors then investigated fixed point outcomes in these
spaces. More information on fixed point theorems in
rectangular metric space is available here, see [3, 4, 5]
Samet et al. [6] developed the concept of α-ψ-
contractive mapping in 2012, which is important be-
cause, unlike the Banach contraction principle, it does
not require the contractive requirements to hold for
every pair of points in the domain. It also takes into
account the scenario of discontinuous mappings. As a
result of these factors, there has been a tremendous in-
crease in the literature dealing with fixed point prob-
lems using admissible mappings (see in [7, 8, 9]).
Most recently, two different generalizations of ad-
missible mapping were given in which the author
Ansari [7] used the idea of C-class functions, whereas
Budhia et al. [11] used a rectangular metric. In
this paper, we prove coincidence and common fixed
point theorems for two mappings in complete Haus-
dorff generalized metric spaces that meet a gener-
alized (ϑ, ψ)-weakly contractive condition. Many
known results in the literature are extended and gen-
eralized by the presented theorems.
2 Preliminaries
Definition 2.1. [2] Let Φ=∅be a set. A generalized
metric (rectangular metric) is a function µ: Φ×Φ→
[0,∞),where the following conditions are fulfilled for
all v, λ, α, δ ∈Φwith α=δand α, δ /∈ {v, λ}:
(i)µ(v, λ) = 0 if and only if v=λ;
(ii)µ(v, λ) = µ(λ, v);
(iii)µ(v, λ)≤µ(v, α) + µ(α, δ) + µ(δ, λ)(quadri-
lateral inequality).
The pair (Φ, µ)is named as a generalized metric
space (a rectangular metric space)
Definition 2.2. [2] Let (Φ, µ)be a rectangular metric
space, and let {vn}be a sequence in Φ.
(i)If (vn, v)→0as n→ ∞,then {vn}is called
rectangular metric space convergent to a limit v.
(ii)If for every ϵ > 0,there exists n(ϵ)∈Nsuch
that µ(vi, vj)< ϵ for all i>j>n(ϵ),then
{vn}is called a rectangular metric space Cauchy
sequence in rectangular metric space.
(iii)A rectangular metric space (Φ, µ)is called com-
plete if every rectangular metric space Cauchy
sequence is rectangular metric space convergent.
Definition 2.3. [7] AC-function F: [0,∞)×
[0,∞)→Ris a continuous function such that for
all v, λ ∈[0,∞):
(i)F(v, λ)≤v;
(ii)F(v, λ) = vimplies that either v= 0 or λ= 0.
1CHUANPIT MUNGKALA, 1ANANTACHAI PADCHAROEN, 2PAKEETA SUKPRASERT
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Volume 21, 2022
1chuanpit.t@rbru.ac.th, 2pakeeta_s@rmutt.ac.th
The letter Cwill denote the class of all C-functions.
Definition 2.4. [8] Let ω, κ : Φ ×Φ→[0,∞)
be two mappings. A map M: Φ →Φis said to
be ω-admissible with respect to κif ω(Mv, Mλ)>
κ(Mv, Mλ)whenever ω(v, λ)> κ(v, λ)for all
v, λ ∈Φ.If κ(v, λ)=1for all v, λ ∈Φ,then M
is called an ω-admissible mapping.
Definition 2.5. [10] A nondecreasing continuous
function ϑ: [0,∞)→[0,∞)is called an altering
distance function if ϑ(t) = 0 if and only if t= 0.
We denote by Ψthe class of altering distance func-
tions.
Lemma 2.6. [11] Let (Φ, µ)be a complete rect-
angular metric space and {vn}be a sequence
in Φsuch that limn→∞ µ(vn, vn+1) = 0 =
limn0∞µ(vn, vn+2)and vn=vmfor all positive in-
tegers n=m. If {vn}is not a Cauchy sequence,
then there exist an ϵ > 0and sequences {mk}and
{nk}in Nwith mk> nk> k with µ(vmk, vnk)>
ϵ, µ(vmk−1, vnk)< ϵ so that the following hold:
(i)limk→∞ µ(vmk−1, vnk+1) = ϵ;
(ii)limk→∞ µ(vmk, vnk) = ϵ;
(iii)limk→∞ µ(vmk−1, vnk) = ϵ;
(iv)limk→∞ µ(vmk, vnk−1) = ϵ;
(v)limk→∞ µ(vmk+1, vnk+1) = ϵ;
Definition 2.7. [3] Let gand Mbe self-mappings of
a nonempty set.
(i)A point ξ∈Φis said to be a common fixed point
of gand Mif ξ=gξ =Mξ.
(ii)A point ξ∈Φis called a coincidence point of g
and Mif gξ =Mξ. And if η=gξ =Mξ, then η
is said to be a point of coincidence of gand M.
(iii)The mappings g, M: Φ →Φare said to be
weakly compatible if they commute at their co-
incidence point that is, gMξ=Mgξ whenever
gξ =Mξ.
Lemma 2.8. [3] Let Φbe a nonempty set. Suppose
that the mappings g, M: Φ →Φhave a unique co-
incidence point ϱin Φ. If gand Mare weakly com-
patible, then gand Mhave a unique common fixed
point.
3 Main Result
Theorem 3.1. Let (Φ, µ)be a Hausdorff rectangular
metric space and M: Φ →Φbe an ω-admissible
mapping with respect to κand let g, M: Φ →Φbe
two self maps such that M(Φ) ⊆g(Φ) and (gΦ, µ)is
a complete rectangular metric space. Suppose there
exist F∈Cand ϑ, ψ ∈Ψsuch that, for v, λ ∈Φ,
ω(v, λ)> κ(v, λ)⇒
ϑ(µ(Mv, Mλ)) ≤F(ϑ(∆(v, λ)), ψ(∆(v, λ))),(1)
where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
Assume that
(i)the pair (g, M)is ω-admissible regarding to the
function κ;
(ii)there exists v0∈Φso that ω(v0, gv0)≥
κ(v0, gv0)and ω(v0,Mv0)≥κ(v0,Mv0);
(iii)gand Mare continuous.
Then gand Mhave a unique coincidence point in Φ.
Moreover, if gand Mare weakly compatible, then g
and Mhave a unique common fixed point.
Proof. First, we shall show the existence of gand M
coincidence point. By induction we get
Consider v0is an arbitrary point. Since M(Φ) ⊆
g(Φ),we create two iterative sequences in Φ,{vn}
and {λn}, as follows:
λn=gvn+1 =Mvn,for all n= 0,1,2, . . . . (2)
If λk=λk−1for some k∈N,then gvk=λk=
λk−1=Mvkand gand Mare have a point of coinci-
dence.
Assume additionally that λn=λn−1for every
n∈N.By letting v=vn, λ =vn+1 into (1) and
condition (2), we get
ω(vn, vn+1)> κ(vn, vn+1)(3)
and
ϑ(µ(λn, λn+1))
=ϑ(µ(Mvn,Mvn+1))
≤F(ϑ(∆(vn, vn+1)), ψ(∆(vn, vn+1))),
(4)
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where
∆(vn, vn+1)
=max {µ(gvn, gvn+1),
µ(gvn,Mvn) + µ(gvn+1,Mvn+1)
2,
µ(gvn,Mvn) + µ(gvn+1,Mvn)
2,
µ(gvn+1,Mvn+1)(1 + µ(gvn,Mvn))
1 + µ(gvn, gvn+1),
µ(gvn,Mvn)(1 + µ(gvn+1,Mvn+1))
1 + µ(Mvn,Mvn+1)}
=max {µ(λn−1, λn),µ(λn−1, λn) + µ(λn, λn+1)
2,
µ(λn−1, λn) + µ(λn, λn)
2,
µ(λn, λn+1)(1 + µ(λn−1, λn))
1 + µ(λn−1, λn)
µ(λn−1, λn)(1 + µ(λn, λn+1))
1 + µ(λn, λn+1)}
=max{µ(λn−1, λn), µ(λn, λn+1)}.
If ∆(vn, vn+1) = µ(λn, λn+1)for some n∈N,then
from (4),
ϑ(µ(λn, λn+1))
=ϑ(µ(Mvn,Mvn+1))
≤F(ϑ(µ(λn, λn+1)), ψ(µ(λn, λn+1)))
≤ϑ(µ(λn, λn+1)).
(5)
Using Definition 2.3, ϑ(µ(λn, λn+1)) = 0 or
ψ(µ(λn, λn+1)) = 0.So µ(λn, λn+1) = 0,which
is a contradiction. Consequently, ∆(vn, vn+1) =
µ(λn−1, λn)for every n∈N.From (4) we get
ϑ(µ(λn, λn+1))
≤F(ϑ(µ(λn−1, λn)), ψ(µ(λn−1, λn)))
≤ϑ(µ(λn−1, λn)).
(6)
Since ϑis nondecreasing,
µ(λn, λn+1)≤µ(λn−1, λn).(7)
Thus, {µ(λn, λn+1)}is a nonincreasing sequence of
positive real numbers, so there exists ϕ≥0such that
the limit
lim
n→∞
µ(λn, λn+1) = ϕ.
Also,
lim
n→∞
µ(λn−1, λn) = ϕ.
From F, ϑ and ψare continuous, we have
lim
n→∞
ϑ(µ(λn, λn+1))
≤lim
n→∞
F(ϑ(µ(λn−1, λn)), ψ(µ(λn−1, λn)))
=F(lim
n→∞
ϑ(µ(λn−1, λn)),lim
n→∞
ψ(µ(λn−1, λn))).
Hence,
ϑ(ϕ)≤F(ϑ(ϕ), ψ(µ(ϕ))) ≤ϑ(ϕ).
Again, using Definition 2.3 we get ϕ= 0,that is,
lim
n→∞
µ(λn, λn+1) = 0.
Now we will show whether µ(λn, λn+2)→0as n→
∞. Utilizing (7) we have
ϑ(µ(λn, λn+2))
=ϑ(µ(Mvn,Mvn+2))
≤F(ϑ(∆(vn, vn+2)), ψ(∆(vn, vn+2)))
≤ϑ(∆(vn, vn+2)).
(8)
Hence,
ϑ(µ(λn, λn+2)) ≤ϑ(∆(vn, vn+2)).
Since ϑis an altering distance, we have
µ(λn, λn+2)≤∆(vn, vn+2).
where
∆(vn, vn+2)
=max {µ(gvn, gvn+2),
µ(gvn,Mvn) + µ(gvn+2,Mvn+2)
2,
µ(gvn,Mvn) + µ(gvn+2,Mvn)
2,
µ(gvn+2,Mvn+2)(1 + µ(gvn,Mvn))
1 + µ(gvn, gvn+2)
µ(gvn,Mvn)(1 + µ(gvn+2,Mvn+2))
1 + µ(Mvn,Mvn+2)}
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=max {µ(λn−1, λn+1),
µ(λn−1, λn) + µ(λn+1, λn+2)
2,
µ(λn−1, λn) + µ(λn+1, λn)
2,
µ(λn+1, λn+2)(1 + µ(λn−1, λn))
1 + µ(λn−1, λn+1)
µ(λn−1, λn)(1 + µ(λn+1, λn+2))
1 + µ(λn, λn+2)}
≤max {µ(λn−1, λn+1),
µ(λn−1, λn) + µ(λn+1, λn+2),
µ(λn−1, λn) + µ(λn+1, λn),
µ(λn+1, λn+2)(1 + µ(λn−1, λn))
µ(λn−1, λn)(1 + µ(λn+1, λn+2))}.
That will be seen limn→∞ ∆(vn, vn+2) =
limn→∞ µ(λn−1, λn+1).From 8 and taking n→ ∞,
we obtain
ϑ(lim
n→∞
µ(λn, λn+2))
≤F(ϑ(lim
n→∞
µ(λn−1, λn+1)), ψ(lim
n→∞
µ(λn−1, λn+1)))
≤ϑ(lim
n→∞
µ(λn−1, λn+1)).
(9)
Hence, the sequence {µ(λn, λn+2)}is non-increasing
and bounded below. Therefore, the sequence
{µ(λn, λn+2)}converges to a number, ϖ≥0.Tak-
ing limit as n→ ∞ in (8), we get
ϑ(ϖ)≤F(ϑ(ϖ), ψ(µ(ϖ))) ≤ϑ(ϖ).
Using Definition 2.3 we get ϖ= 0,that is,
lim
n→∞
µ(λn, λn+2) = 0.(10)
Assume that λn=λmfor all m=nand demonstrate
that {λn}is an rectangular metric spaces Cauchy se-
quence. If feasible, make {λn}not a Cauchy se-
quence, according to Lemma 2.6, there exists ϵ > 0
such that we may identify the subsequences {λmk}
and {λnk}of {λn}with mk> nk> k such that
lim
k→∞
µ(λmk, λnk) = lim
k→∞
µ(λmk−1, λnk−1) = ϵ.
(11)
We now substitute v=vnkand λ=vmkin (1). Con-
sider
ϑ(λnk, λmk)
=ϑ(µ(Mvnk,Mvmk))
≤F(ϑ(∆(vnk, vmk)), ψ(∆(vnk, vmk))),
(12)
where
∆(vnk, vmk)
=max {µ(gvnk, gvmk),
µ(gvnk,Mvnk) + µ(gvmk,Mvmk)
2,
µ(gvnk,Mvnk) + µ(gvmk,Mvnk)
2,
µ(gvmk,Mvmk)(1 + µ(gvnk,Mvnk))
1 + µ(gvnk, gvmk),
µ(gvnk,Mvnk)(1 + µ(gvmk,Mvmk))
1 + µ(Mvnk,Mvmk)}
=max {µ(λnk−1, λmk−1),
µ(λnk−1, λnk) + µ(λmk−1, λmk)
2,
µ(λnk−1, λnk) + µ(λmk−1, λnk)
2,
µ(λmk−1, λmk)(1 + µ(λnk−1, λvnk))
1 + µ(λnk−1, λmk−1),
µ(λnk−1, λnk)(1 + µ(λmk−1, λvmk))
1 + µ(λnk, λmk)}.
Then
lim
k→∞
∆(vnk, vmk) = ϵ. (13)
From condition (2), we get
ω(vnk, vmk)≥κ(vnk, vmk).(14)
Using (12) and (12), we get
ϑ(ϵ)≤ F(ϑ(ϵ), ψ(ϵ)) ≤ϑ(ϵ).
This implies that ϑ(ϵ)=0or ψ(ϵ)) = 0,thus, ϵ= 0,
but this is a contradiction with the fact ϵ > 0.Thus,
{λn}is a rectangular metric space Cauchy sequence.
Since (g(Φ),Φ) is rectangular metric space complete,
there exists η∈g(Φ) such that λn→ηas n→ ∞.
Let ξ∈Φbe such that gξ =η. Then
lim
n→∞
λn=gξ. (15)
We shall prove that Mξ=gξ. Applying the inequality
(18), with v=vn,and λ=ξwe obtain
ω(vn, ξ)> κ(vn, ξ).
This implies that
ϑ(µ(λn,Mξ))
=ϑ(µ(Mvn,Mξ))
≤F(ϑ(∆(vn, ξ)), ψ(∆(vn, ξ))),
(16)
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where
∆(vn, ξ)
=max {µ(gvn, gξ),µ(gvn,Mvn) + µ(gξ, Mξ)
2,
µ(gvn,Mvn) + µ(gξ, Mvn)
2,
µ(gξ, Mξ)(1 + µ(gvn,Mvn))
1 + µ(gvn, gξ),
µ(gvn,Mvn)(1 + µ(gξ, Mξ))
1 + µ(Mvn,Mξ)}
=max {µ(λn−1, gξ),µ(λn−1, λn) + µ(gξ, Mξ)
2,
µ(λn−1, λn) + µ(gξ, λn)
2,
µ(gξ, Mξ)(1 + µ(λn−1, λn))
1 + µ(λn−1, gξ),
µ(λn−1, λn)(1 + µ(gξ, Mξ))
1 + µ(λn,Mξ)}.
Since Φis Hausdorff, {λn} → ηwhere n→ ∞, we
deduce that
lim
n→∞ ∆(vn, ξ) = µ(gξ, Mξ).(17)
Taking limit as n→ ∞ in (16),
ϑ(µ(gξ, Mξ))
≤F(ϑ(µ(gξ, Mξ)), ψ(µ(gξ, Mξ)))
≤ϑ(µ(gξ, Mξ).)
Consequently, we get ϑ(µ(gξ, Mξ)) = 0 or
ψ(µ(gξ, Mξ)) = 0 hence µ(gξ, Mξ)) = 0, that is,
gξ =Mξ. Thus we proved that η=gξ =Mξand so
ηis a point of coincidence of gand M.
Now, we show that if the coincidence point of g
and Mexists, it is unique. Let η1and η2be the gand
Mcoincidence points. Thus, there exists some v, λ ∈
Φsuch that η1=Mv=gv and η2=Mλ=gλ. We
can deduce from (1) that
ω(v, λ)> κ(v, λ)⇒
ϑ(µ(η1, η2)) = ϑ(µ(Mv, Mλ))
≤F(ϑ(∆(v, λ)), ψ(∆(v, λ))),
(18)
where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,
µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}
=max {µ(η1, η2),µ(η1, η1) + µ(η2, η2)
2,
µ(η1, η1) + µ(η2, η1)
2,µ(η2, η2)(1 + µ(η1, η1))
1 + µ(η1, η2),
µ(η1, η1)(1 + µ(η2, η2))
1 + µ(η1, η2)}
=µ(η1, η2).
As a result, we deduce that η1=η2by (18).
Recall that gand Mare only weakly compatible.
By Lemma 2.8, the point ηis the unique common
fixed point of gand Msince it is the unique coin-
cidence point of gand M.
Corollary 3.2. Let (Φ, µ)be a Hausdorff rectangular
metric space and M: Φ →Φbe an ω-admissible
mapping with respect to κand let g, M: Φ →Φbe
two self maps such that M(Φ) ⊆g(Φ) and (gΦ, µ)is
a complete rectangular metric space. Suppose there
exist F∈Cand ϑ, ψ ∈Ψsuch that, for v, λ ∈Φ,
ω(v, λ)> κ(v, λ)⇒
ϑ(µ(Mv, Mλ)) ≤ϑ(∆(v, λ)) −ψ(∆(v, λ)),
where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
Assume that
(i)the pair (g, M)is ω-admissible regarding to the
function κ;
(ii)there exists v0∈Φso that ω(v0, gv0)≥
κ(v0, gv0)and ω(v0,Mv0)≥κ(v0,Mv0);
(iii)gand Mare continuous.
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Then gand Mhave a unique coincidence point in Φ.
Moreover, if gand Mare weakly compatible, then g
and Mhave a unique common fixed point.
Proof. Let F(v, λ) = v−λin Theorem 3.1, we have
following corollary 3.2.
Corollary 3.3. Let (Φ, µ)be a Hausdorff rectangular
metric space and M: Φ →Φbe an ω-admissible
mapping with respect to κand let g, M: Φ →Φbe
two self maps such that M(Φ) ⊆g(Φ) and (gΦ, µ)is
a complete rectangular metric space. Suppose ψ∈Ψ
and v, λ ∈Φ,such that
ω(v, λ)> κ(v, λ)⇒
µ(Mv, Mλ)≤∆(v, λ)−ψ(∆(v, λ)),
where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,
µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
Assume that
(i)the pair (g, M)is ω-admissible regarding to the
function κ;
(ii)there exists v0∈Φso that ω(v0, gv0)≥
κ(v0, gv0)and ω(v0,Mv0)≥κ(v0,Mv0);
(iii)gand Mare continuous.
Then gand Mhave a unique coincidence point in Φ.
Moreover, if gand Mare weakly compatible, then g
and Mhave a unique common fixed point.
Proof. Let ϑ(t) = tin Corollary 3.2, we have follow-
ing corollary 3.3.
Corollary 3.4. Let (Φ, µ)be a Hausdorff rectangular
metric space and M: Φ →Φbe an ω-admissible
mapping with respect to κand let g, M: Φ →Φbe
two self maps such that M(Φ) ⊆g(Φ) and (gΦ, µ)is
a complete rectangular metric space. For all v, λ ∈
Φand 0< k < 1such that such that
ω(v, λ)> κ(v, λ)⇒µ(Mv, Mλ)≤k∆(v, λ),
where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,
µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
Assume that
(i)the pair (g, M)is ω-admissible regarding to the
function κ;
(ii)there exists v0∈Φso that ω(v0, gv0)≥
κ(v0, gv0)and ω(v0,Mv0)≥κ(v0,Mv0);
(iii)gand Mare continuous.
Then gand Mhave a unique coincidence point in Φ.
Moreover, if gand Mare weakly compatible, then g
and Mhave a unique common fixed point.
Proof. Let ψ(t) = (1−k)(t)for 0< k < 1in Corol-
lary 3.3, we have following corollary 3.4.
4 Applications
Definition 4.1. Let Υbe the class of functions χ:
[0,∞)→[0,∞)satisfying the following
(1) χis Lebesgue integrable function on each com-
pact subset of [0,∞);
(2) ∫ϵ
0
χ(t)dt > 0for any ϵ > 0.
Theorem 4.2. Let (Φ, µ)be a Hausdorff rectangular
metric space and let g, M: Φ →Φbe two self maps
such that M(Φ) ⊆g(Φ) and (gΦ, µ)is a complete
rectangular metric space and that the following con-
dition holds:
∫µ(Mv,Mλ)
0
χ(t)dt ≤∫∆(v,λ)
0
χ(t)dt−∫∆(v,λ)
0
φ(t)dt
(19)
for all v, λ ∈Φand χ, φ ∈Υ,such that gand M
satisfy inequality (1), where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,
µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2022.21.60
Chuanpit Mungkala, Anantachai Padcharoen, Pakeeta Sukprasert
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545
Then gand Mhave a unique coincidence point.
Proof. Let ϑ(t) = ∫t
0
χ(τ)dτ and ψ(t) =
∫t
0
φ(τ)dτ. Then, ϑ, ψ ∈Ψ. Thus, by Theorem 3.1,
gand Mhave a unique coincide fixed point.
Theorem 4.3. Let (Φ, µ)be a Hausdorff rectangular
metric space and let g, M: Φ →Φbe two self maps
such that M(Φ) ⊆g(Φ) and (gΦ, µ)is a complete
rectangular metric space and that the following con-
dition holds:
∫µ(Mv,Mλ)
0
χ(t)dt ≤κ∫∆(v,λ)
0
χ(t)dt (20)
for all v, λ ∈Φ, χ ∈Υand 0≤κ < 1such that g
and Msatisfy inequality (1), where
∆(v, λ)
=max {µ(gv, gλ),µ(gv, Mv) + µ(gλ, Mλ)
2,
µ(gv, Mv) + µ(gλ, Mv)
2,
µ(gλ, Mλ)(1 + µ(gv, Mv))
1 + µ(gv, gλ),
µ(gv, Mv)(1 + µ(gλ, Mλ))
1 + µ(Mv, Mλ)}.
Then gand Mhave a unique coincidence point.
Proof. Let f(t) = χ(t)−κχ(t)Thus, by Theorem
3.1, gand Mhave a unique coincide fixed point.
5 Acknowledgments
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2022.21.60
Chuanpit Mungkala, Anantachai Padcharoen, Pakeeta Sukprasert
E-ISSN: 2224-2880
546
Volume 21, 2022
Firstly, Chuanpit Mungkala (chuanpit.t@rbru.ac.th)
and Anantachai Padcharoen
(anantachai.p@rbru.ac.th) were financially
supported by the Research and Development
Institute of Rambhaibarni Rajabhat University.
Finally, Pakeeta Sukprasert
(pakeeta\_s@rmutt.ac.th) was financially supported
by Rajamangala University of Technology
Thanyaburi (RMUTT).