Some generalized fixed point theorems in the context of ordered metric spaces

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Research Article

Some generalized fixed point theorems in the context of ordered metric spaces

Mehmet Kir^∗, Hukmi Kiziltunc

Department of Mathematics, Faculty of Science, Ataturk University, 25240, Erzurum, Turkey.

Abstract

In this paper, we give three main theorems which are new generalizations of Banach fixed point theorem, Kannan fixed point theorem and Chatterjea fixed point theorem in the context of the ordered metric space.

Keywords: Fixed point, Chatterjea fixed point theorem, Kannan fixed point theorem, contraction mappings

2010 MSC: 41A65, 41A15, 47H09, 47H10, 54H25.

1. Introduction and Preliminaries

A mappingT :X →X where (X, d) is a metric space, is said to be a contraction if there existsk∈[0,1) such that for allx, y∈X,

d(T x, T y)≤kd(x, y).

Banach proved that a contraction mapping has a unique fixed point in a complete metric space (X, d) [1] (see also [2, 5, 6, 9]).

In [11], Ran and Reurings established the Banach fixed point theorem in the context of ordered metric spaces.

Theorem 1.1([11]). Let (X,≼)be a partially ordered set endowed with a metric dand(X, d) be a complete metric space. Furthermore, every pair x, y∈X has a lower bound and an upper bound. If f :X →X is a continuous, monotone (i.e., either order-preserving or order-reversing) map fromX into X such that

∃ 0< c <1 :d(f x, f y)≤cd(x, y), x≥y

∗Corresponding author

Email addresses: [email protected](Mehmet Kir),[email protected](Hukmi Kiziltunc) Received 2015-1-7

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and

∃x0∈X:x0≤f x0 or x0 ≥f x0, then f has a unique fixed point x. Morever, for every x∈X lim

n→∞fⁿx= x.

Nieto and L´opez [10] gave an alternative condition for the continuity of the mapping f as following;

”if any nondecreasing sequence{xn}inXconverges to z thenxn≼z for all n≥0”

Also, to guarantee the uniqueness of the fixed point, Nieto and L´opez [10] gave an alternative condition met the requirement of ” every pairx, y∈X has a lower bound and an upper bound” as following;

”for every x, y∈X; there exists z∈X which is comparable tox and y ”. (1.1) Theorem 1.2 ([10]). Let(X,≼)be an ordered set endowed with a metric dand the mapping f :X →X be given. Suppose that the following conditions hold:

i) (X, d) is complete;

ii) if any nondecreasing sequence {xn} in X converges toz, then xn≼z for alln≥0;

iii) f is nondecreasing;

iv) there exists x₀ ∈X such that x₀ ≼T x₀;

v) there exists a constant k∈(0,1) such that for all x, y∈X withx≽y, d(f x, f y)≤kd(x, y).

Then f has a fixed point. Moreover, if for all(x, y)∈X×X there exists a z∈X such that x≼z and y≼z, then the fixed point is unique.

Two of the well known fixed point theorems are Kannan fixed point theorem and Chatterjea fixed point theorem.

Theorem 1.3 ([4]). If a mapping T : X → X where (X, d) is a complete metric space, satisfies the inequality

d(T x, T y)≤a[d(x, T x) +d(y, T y)] (1.2) where a∈ [

0,¹₂)

and x, y ∈ X, then T has a unique fixed point. The mappings satisfying (1.2) are called Kannan type mappings.

Theorem 1.4([3]). If a mappingT :X →Xwhere(X, d)is a complete metric space, satisfies the inequality d(T x, T y)≤b[d(x, T y) +d(y, T x)]

such thatb∈[ 0,¹₂)

and x, y∈X, then T has a unique fixed point.

Also, in 2011, Moradi and Davood introduced a new extension of Kannan fixed point theorem on complete metric space as following;

Theorem 1.5 ([8]). Let (X, d) be a complete metric space and T, S :X → X be mappings such that T is continuous, one to one and subsequentially convergent. If µ∈[

0,¹₂)

and x, y∈X, d(T Sx, T Sy)≤µ[d(T x, T Sy) +d(T y, T Sx)],

then,S has a unique fixed point. Also, if T is sequentially convergent then for everyx₀ ∈X the sequence of iterates {Sⁿx0} converges to the fixed point.

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Definition 1.6 ([8]). Let (X, d) be a metric space.

i) A mapping T :X → X is said to be sequentially convergent if we have, for every sequence {yn}, if {T yn} is convergence then{yn}is also convergence.

ii)T is said to be subsequentially convergent if we have, for every sequence{y_n}, if{T y_n}is convergence then{yn}has a convergent subsequence.

In 2014, Mustafa et. al., [7] introduced fixed point theorems for weakly T-Chatterjea and weakly T-Kannan-contractive mappings in completeb−metric spaces.

In this paper we have three main theorems which are new generalization of Banach fixed point theorem, Kannan fixed point theorem and Chatterjea fixed point theorem in the context of the ordered metric space.

2. Main Results

For the simplicity in writing, we will use the following symbols. We donote by Ψ the set of all functions F : [0,∞)→[0,∞) satisfying:

F1)F is continuous and monotone nondecreasing, F2)F(t) = 0 if and only if t= 0.

Also, we denote bySSC(X) the set of all mappingsT :X →X such thatT is one to one, continuous, subsequentially convergent and preserve the order, bySC(X) the set of all mappingsT :X →X such that T is one to one, continuous, sequentially convergent and preserve the order.

Theorem 2.1. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X), F ∈Ψ. For all x, y∈X withx≼y, α ∈[0,1),

F(d(T f x, T f y))≤αF(d(T x, T y)). Also, suppose that either

C1) f is continuous or C2) Assume that if any nondecreasing sequence{x_n} in X converges to z, then xn≼z for alln≥0.

If there exists x0 ∈X withx0 ≼f x0,thenf has a fixed point inX. Moreover, if for each x, y∈X there exists z∈X which is comparable to x and y, then the fixed point is unique.

Proof. Letx0 ∈Xbe an arbitrary point such thatxn=f xn−1 =fⁿx0,n= 1,2,3,· · ·. Asfis nondecreasing and x₀ ≼f x₀, we have

T x0≼T f x0 ≼T f²x0 ≼T f³x0 ≼ · · · ≼T fⁿx0≼ · · · (2.1) Since T x_n≼T x_n+1,

F(d(T x_n, T x_n+1)) = F(d(T f x_n₋₁, T f x_n))

≤ αF(d(T x_n₋₁, T x_n)) ...

≤ αⁿF(d(T x0, T x1)) . (2.2)

Lettingn→ ∞in (2.2), then we have

F(d(T xn, T xn+1))→0⁺, asn→ ∞.

d(T xn, T xn+1)→0, as n→ ∞.

Also, form, n∈N, m > n

F(d(T x_n, T x_m))≤αⁿF(d(T x₀, T x_m₋_n)) (2.3)

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letting m, n→ ∞(2.3), we get

d(T xn, T xm)→0.

Thus, we obtain that{T x_n} is Cauchy sequence. As (X, d) is complete, there exists v∈X such that

nlim→∞T x_n=v. (2.4)

Note thatT ∈SSC(X), then {x_n} has a convergent subsequence, so there isu∈Xsuch that

klim→∞ x_n(k)=u. (2.5)

Also,T is continuous andx_n(k)→u,therefore

k→∞limT x_n(k) =T u. (2.6)

and

klim→∞d(

T x_n(k), T u)

= 0.

Now, we will show that u∈X is a fixed point off. In here we have two cases.

Case 1: Let C1) holds. From the continuity of f, we have T u= lim

k→∞T x_n(k)= lim

k→∞T f x_n(k)₋₁ =T f u.

since,T is one to one we get f u=u, namely u∈X is a fixed point off.

Case 2: Let C2) holds. Since {

T x_n(k)}

converges toT u∈X,for all ^ϵ₂ >0 there isN1 ∈N such that for all n(k)> N₁, we have

d(

T x_n(k), T u)

< ϵ 2. Also, as{

T x_n(k)}

converges to T u.From C2), we get T x_n(k) ≼T u and we have F

( d

(

T f^n(k)+1x, T f u

))≤αF (

d (

T f^n(k)x, T u ))

. (2.7)

Lettingk→ ∞ in (2.7), we have

F(d(T u, T f u))≤0 (2.8)

The inequality (2.8) implies that T u=T f u.As,Tis one to one we get u∈X is a fixed point off.

Adding the condition (1.1), we show the uniqueness of the fixed point. We will do this by showing that

klim→∞f^n(k)x=u for everyx∈X. Here we have two cases.

Firstly, let x andx₀ be comparable, thenx≼x₀ orx₀≼x. In both cases, we have f^n(k)x≼f^n(k)x₀ or f^n(k)x0 ≼f^n(k)x.Hence, we have

F (

d (

T f^n(k)x, T f^n(k)x₀

))≤α^n(k)F(d(T x, T x₀)) letting k→ ∞in the last inequality, we have

klim→∞f^n(k)x= lim

k→∞f^n(k)x0=u.

Secondly, letxandx0 are not comparable and letx1,resp. x2,be an upper bound, resp. a lower bound, ofx and x₀.That is x₁ ≽x≽x₂ and x₁ ≽x₀≽x₂. Thus we have

klim→∞f^n(k)x= lim

k→∞f^n(k)x₀=u.

Hence, the proof is completed.

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Remark 2.2. The case of T ∈ SC(X), the proof of the Theorem 2.1 takes place analogously by replacing {n}with {n(k)}.

Corollary 2.3. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, α ∈[0,1),

d(T f x, T f y)≤αd(T x, T y).

Also, suppose that the condition C1) or C2) holds. If there exists x₀ ∈X withx₀≼f x₀,thenf has a fixed point in X. Moreover, if for eachx, y∈X there exists z∈X which is comparable to x andy, then the fixed point is unique.

Corollary 2.4. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be a monotone nondecreasing mapping and F ∈ Ψ. For all x, y ∈ X with x≼y, α∈[0,1),

F(d(f x, f y))≤αF(d(x, y)).

Also, suppose that the condition C1) or C2) holds. If there exists x0 ∈X withx0≼f x0,thenf has a fixed point in X. Moreover, if for eachx, y∈X there exists z∈X which is comparable to x andy, then the fixed point is unique.

Corollary 2.5. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be mappings such that f is monotone nondecreasing. For all x, y ∈X with x≼y, α∈[0,1),

d(f x, f y)≤αd(x, y).

Also, suppose that the condition C1) or C2) holds. If there exists x0 ∈X withx0≼f x0,thenf has a fixed point in X. Moreover, if for eachx, y∈X there exists z∈X which is comparable to x andy, then the fixed point is unique.

Corollary 2.6. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, α ∈[0,1), ∫ d(T f x,T f y)

0

φ(t)dt≤α

∫ _{d(T x,T y)}

0

φ(t)dt, (2.9)

where φ: [0,∞)→ [0,∞) is a Lebesgue-integrable mapping which summeble ( i.e., with finite integral ) on each compact subset of [0,∞), nonnegative, and such that for each ϵ >0,∫_ϵ

0 φ(t)dt >0.

Also, suppose that the condition C1) or C2) holds. If there exists x0 ∈X withx0 ≼f x0, then f has a fixed point inX. Moreover, if for each x, y∈X there exists z∈X which is comparable to x andy, then the fixed point is unique.

Corollary 2.7. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be mappings such that f is monotone nondecreasing. For all x, y ∈X with x≼y, α∈[0,1), ∫ _{d(f x,f y)}

0

φ(t)dt≤α

∫ _d(x,y)

0

φ(t)dt, (2.10)

0 φ(t)dt >0.

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Also, suppose that the condition C1) or C2) holds. If there exists x₀ ∈X withx₀ ≼f x₀, then f has a fixed point inX. Moreover, if for each x, y∈X there exists z∈X which is comparable to x andy, then the fixed point is unique.

Theorem 2.8. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X), F ∈Ψ. For all x, y∈X withx≼y, β ∈[

0,¹₂) ,

F(d(T f x, T f y))≤β[F(d(T x, T f x)) +F(d(T y, T f y))]. Also, suppose that

If there exists x₀ ∈X withx₀ ≼f x₀,thenf has a fixed point inX. Moreover, if for each x, y∈X there exists z∈X which is comparable to x and y, then the fixed point is unique.

Proof. Letx₀∈Xbe an arbitrary point such thatx_n=f x_n₋₁ =fⁿx₀,n= 1,2,3, .... Asf is nondecreasing and x₀ ≼f x₀, we have

T x0≼T f x0 ≼T f²x0 ≼T f³x0 ≼ · · · ≼T fⁿx0≼ · · · (2.11) Since T xn≼T xn+1

≤ β[F(d(T xn−1, T xn)) +F(d(T xn, T xn+1))]. (2.12) From, (2.12), we get

F(d(T x_n, T x_n+1))≤ β

1−βF(d(T x_n₋₁, T x_n)). (2.13) Also, by continuing the process (2.13), we obtain that

F(d(T x_n, T x_n+1))≤ ( β

1−β )n

F(d(T x₀, T x₁)) . (2.14) Lettingn→ ∞in (2.14), we obtain that

F(d(T xn, T xn+1))→0⁺ asn→ ∞. Again using (2.14), for allm, n∈N, taking m > n,we have

F(d(T xn, T xm)) = F(d(T fⁿx0, T f^mx0))

≤

( β 1−β

)n

F( d(

T x₀, T f^m⁻ⁿx₀))

. (2.15)

Lettingm, n→ ∞ in (2.15) ,we have

F(d(T x_n, T x_m))→0⁺asm, n→ ∞. So, we haved(T xn, T xm)→0 as, m, n→ ∞.

In the next stage, by using similar methods in Theorem 2.1, we obtain that{T x_n} is Cauchy sequence in complete metric space (X, d) and there existu∈X such that{T x_n} converges toT u∈X. Now, we will show thatu∈X is a fixed point off. In here we have two cases.

Case 1: Let C1) holds. From the continuity of f, we have T u= lim

k→∞T x_n(k)= lim

k→∞T f x_n(k)₋₁ =T f u.

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Since,T is one to one we get that u∈X is a fixed point off.

Case 2: Let C2) holds. Since {T xn}converges toT u∈X,for all ₂^ϵ >0 there isN1 ∈Nsuch that for all n > N1,we have

d(T xn, T u)< ϵ 2. Also, as{T x_n} converges toT u,From C2) we getT x_n≼T u and

F(d(T u, T f u)) ≤ F(d(T u, T xn) +d(T xn, T f u))

= F(d(T u, T xn) +d(T fⁿx0, T f u))

≤ F(d(T u, T x_n) +d(T fⁿx₀, T fⁿx₁) +d(T fⁿx₁, T f u))

= F(d(T u, T x_n) +d(T f x_n, T f x_n+1) +d(T f x_n, T f u)). (2.16) Lettingn→ ∞in (2.7), we have

F(d(T u, T f u))≤0 (2.17)

The inequality (2.17) implies thatT u=T f u.As, T is one to one we getu∈X is a fixed point off.Adding the condition (1.1), we show the uniqueness of the fixed point.

Let u^′ ∈ X be another fixed point of f. From the (1.1), there exists an element z in X such that z comparable tou and u^′.The monotonicity of f implies thatfⁿz comparable to fⁿu=u andfⁿu^′ =u^′ for all n∈N.As, T ∈SSC(X) , then T fⁿzis comparable to T u and T u^′. Also,F ∈Ψ we have

F( d(

T u, T u^′))

≤ F(

d(T u, T fⁿz) +d(

T fⁿz, T u^′))

= F(d(T fⁿu, T fⁿz)) +F( d(

T fⁿz, T fⁿu^′))

≤ βⁿF(d(u, z)) +βⁿF( d(

z, u^′))

. (2.18)

Lettingn→ ∞in the inequality (2.18), F(

d(

T u, T u^′))

≤0.

The last inequality impliesT u=T u^′.AsT is one to one we getu=u^′.Hence, the proof is completed.

Remark 2.9. The case of T ∈SC(X), the proof of the Theorem 2.8 takes place analogously by replacing {n}with {n(k)}.

Corollary 2.10. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, β ∈[

0,¹₂) ,

d(T f x, T f y)≤β[d(T x, T f x) +d(T y, T f y)]. Also, suppose that

C1) f is continuous or C2) Assume that if any nondecreasing sequence{xn} in X converges to z, then x_n≼z for alln≥0.

Corollary 2.11. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be a monotone nondecreasing mapping and F ∈ Ψ. For all x, y ∈ X with x≼y, β ∈[

0,¹₂) ,

F(d(f x, f y))≤β[F(d(x, f x)) +F(d(y, f y))].

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Also, suppose that C1) f is continuous or

C2) Assume that if any nondecreasing sequence {xn} in X converges toz, then xn≼z for alln≥0.

Corollary 2.12. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Letf :X →Xbe a monotone nondecreasing mapping. For allx, y∈Xwithx≼y, β∈[

0,¹₂) ,

d(f x, f y)≤β[d(x, f x) +d(y, f y)]. Also, suppose that

C1)f is continuous or

C2) Assume that if any nondecreasing sequence{x_n} in X converges to z, then x_n≼z for all n≥0.

Remark 2.13. The Corollary 2.12 is Kannan fixed point theorem in the context of ordered metric space.

Corollary 2.14. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, β ∈[

0,¹₂)

, ∫ d(T f x,T f y) 0

φ(t)dt≤β

∫ d(T x,T f x)+d(T y,T f y) 0

φ(t)dt, (2.19)

0 φ(t)dt >0.

Corollary 2.15. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Letf :X →Xbe a monotone nondecreasing mapping. For allx, y∈Xwithx≼y, β∈[

0,¹₂) ,

∫ _{d(f x,f y)}

0

φ(t)dt≤β

∫ d(x,f x)+d(y,f y) 0

φ(t)dt, (2.20)

0 φ(t)dt >0.

Remark 2.16. Also, by taking φ(t) = 1 in Corollary 2.15, we obtain Kannan fixed point theorem in the context of the ordered metric space.

Theorem 2.17. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be a monotone nondecreasing mapping and T ∈ SSC(X), F ∈ Ψ. For all x, y∈X withx≼y, µ∈[

0,¹₂) ,

F(d(T f x, T f y))≤µ[F(d(T x, T f y)) +F(d(T y, T f x))]. Also, suppose that

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Proof. Letx₀∈Xbe an arbitrary point such thatx_n=f x_n₋₁ =fⁿx₀,n= 1,2,3, .... Asf is nondecreasing and x0 ≼f x0, we have

T x0≼T f x0 ≼T f²x0 ≼T f³x0 ≼ · · · ≼T fⁿx0≼ · · · (2.21) ConsideringF ∈Ψ andT xn≼T xn+1

≤ µ[F(d(T x_n−1, T x_n+1)) +F(d(T x_n, T x_n))]

≤ µ[F(d(T xn−1, T xn)) +F(d(T xn, T xn+1))]. (2.22) From inequality (2.22), we have

F(d(T x_n, T x_n+1))≤ µ

1−µF(d(T x_n₋₁, T x_n)), and we get

F(d(T xn, T xn+1))≤ ( µ

1−µ )n

F(d(T x0, T x1)). (2.23) Lettingn→ ∞in (2.23), we obtain that

F(d(T xn, T xn+1))→0⁺, ( as n→ ∞ ).

Also, using (2.23), for allm, n∈N, taking m > n,we have F(d(T x_n, T x_m)) = F(d(T fⁿx₀, T f^mx₀))

≤

( µ 1−µ

)_n F(

d(

T x0, T f^m⁻ⁿx0

)) (2.24)

Lettingm, n→ ∞ in (2.24), we have

F(d(T x_n, T x_m))→0⁺ asm, n→ ∞.

The last inequality implies d(T xn, T xm) → 0 as, m, n → ∞. we obtain that {T xn} is Cauchy sequence in complete metric space (X, d) and there existu ∈X such that {T x_n} converges to T u∈X. In the next stage, by using similar methods in Theorem 2.1 or Theorem 2.8, the proof can be completed.

Corollary 2.18. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, µ∈[

0,¹₂) ,

d(T f x, T f y)≤µ[d(T x, T f y) +d(T y, T f x)]. Also, suppose that

C1) f is continuous or C2) assume that if any nondecreasing sequence {x_n} in X converges toz, then x_n≼z for alln≥0.

Corollary 2.19. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f : X → X be a monotone nondecreasing mapping and F ∈ Ψ. For all x, y ∈ X with x≼y, µ∈[

0,¹₂) ,

F(d(f x, f y))≤µ[F(d(x, f y)) +F(d(y, f x))].

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Also, suppose that

C1) f is continuous or C2) Assume that if any nondecreasing sequence{xn} in X converges to z, then xn≼z for alln≥0.

If there exists x₀ ∈X with x₀ ≼f x₀, thenf has a fixed point in X. Moreover, if for all(x, y)∈X×X there exists a z∈X such thatx≼z andy ≼z, then the fixed point is unique.

Corollary 2.20. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Letf :X →Xbe a monotone nondecreasing mapping. For allx, y∈Xwithx≼y, µ∈[

0,¹₂) ,

d(f x, f y)≤µ[d(x, f y) +d(y, f x)]. Also, suppose that

If there exists x₀ ∈X with x₀ ≼f x₀,then f has a fixed point in X.Moreover, if for eachx, y∈X there exists z∈X which is comparable to x and y, then the fixed point is unique.

Remark 2.21. The Corollary 2.20 is Chatterjea fixed point theorem in the context of ordered metric space.

Corollary 2.22. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Let f :X → X be a monotone nondecreasing mapping and T ∈SSC(X). For all x, y ∈X withx≼y, µ∈[

0,¹₂)

, ∫ d(T f x,T f y) 0

φ(t)dt≤µ

∫ d(T x,T f y)+d(T y,T f x) 0

φ(t)dt, (2.25)

0 φ(t)dt >0.

Corollary 2.23. Let (X,≼) be a partially ordered set endowed with a metric d and (X, d) be a complete metric space. Letf :X →Xbe a monotone nondecreasing mapping. For allx, y∈Xwithx≼y, µ∈[

0,¹₂) ,

∫ _{d(f x,f y)}

0

φ(t)dt≤µ

∫ d(x,f y)+d(y,f x) 0

φ(t)dt, (2.26)

where φ: [0,∞)→ [0,∞) is a Lebesgue-integrable mapping which summeble ( i.e., with finite integral ) on each compact subset of [0,∞), nonnegative, and such that for each ϵ >0,∫_ϵ

0 φ(t)dt >0.

Remark 2.24. Also, takingφ(t) = 1 in Corollary 2.23, we get Chatterjea fixed point theorem in the context of ordered metric space.

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