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(1)RAN-REURINGS FIXED POINT THEOREM IS AN IMMEDIATE CONSEQUENCE OF THE BANACH CONTRACTION PRINCIPLE BESSEM SAMET Abstract

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RAN-REURINGS FIXED POINT THEOREM IS AN IMMEDIATE CONSEQUENCE OF THE BANACH CONTRACTION PRINCIPLE

BESSEM SAMET

Abstract. In this short note, we prove in few lines that Ran-Reurings fixed point theorem [Proc.

Amer. Math. Soc.132 (2004) 1435–1443] is an immediate consequence of the famous Banach contraction principle.

1. Introduction

In order to establish the existence of a positive solution to a certain nonlinear matrix equation, Ran and Reurings [2] established a fixed point theorem in a metric space endowed with a partial order. After this work, this theorem was extended and generalized in many directions. In this note, we prove that Ran-Reurings fixed point theorem is an immediate consequence of the well-known Banach fixed point theorem [1].

Let us recall the Banach contraction principle.

Theorem 1.1. Let (Z, d)be a complete metric space and T :Z → Z be a given mapping. Suppose that there exists some constantλ∈(0,1) such that

d(T x, T y)≤λd(x, y), (x, y)∈ Z × Z.

ThenT has a unique fixed.

LetX be a nonempty set endowed with a partial order .

Definition 1.2. We say thatT :X→X is non-decreasing with respect to the partial orderif (x, y)∈X×X, xy=⇒T xT y.

Ran and Reurings [2] established the following result.

Theorem 1.3. Let(X, d)be a complete metric space endowed with a partial order. LetT :X→X be a continuous and non-decreasing mapping with respect to . Suppose that there exists some constant λ∈(0,1) such that

d(T x, T y)≤λd(x, y), (x, y)∈X×X, xy.

If there exists somex0∈X such that x0T x0, thenT has a fixed point.

2. Discussion Now, we prove that

Theorem1.1=⇒ Theorem1.3.

Let us consider the subset ΛT(x0) ofX defined by

ΛT(x0) ={Tnx0: n= 0,1,2,· · · }.

2010Mathematics Subject Classification. 47H10.

Key words and phrases. Fixed point, Ran-Reurings fixed point theorem, Banach contraction principle.

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2 B. SAMET

Let

Z= ΛT(x0)

be the closure of ΛT(x0) with respect to the metric d. Clearly, (Z, d) is a complete metric space.

We claim that

T(Z)⊆ Z.

Let z ∈ Z. From the definition of Z, there exists a sequence {Tnkx0}k that converges toz with respect to the metricd. The continuity of T yields{Tnk+1x0}k converges toT z with respect to the metricd. Since{Tnk+1x0}k⊆ Z andZ is closed, then T z∈ Z, which proves our claim.

Now, let (x, y) be an arbitrary pair of points inZ × Z. From the definition ofZ, there exists a sequence{Tnkx0}kthat converges toxwith respect to the metricd. Similarly, there exists a sequence {Tnpx0}p that converges toy with respect to the metricd. On the other hand, the monotony ofT yields

x0T x0T2x0 · · · Tnx0Tn+1x0 · · ·

ThenTnkx0andTnpx0are comparable with respect to the partial orderfor every natural numbers pandk. Thus we have

d(Tnk+1x0, Tnp+1x0)≤λd(Tnkx0, Tnpx0), for allk, p.

Lettingk→ ∞ andp→ ∞in the above inequality, using the continuity ofT and the metricd, we obtain

d(T x, T y)≤λd(x, y).

As consequence, we have

d(T x, T y)≤λd(x, y), (x, y)∈ Z × Z.

Finally, the Banach contraction principle (Theorem 1.1) gives us the existence of a unique fixed point ofT in Z. Note that the uniqueness is obtained just in the subspace Z of X. So,T has at least one fixed point in the hole spaceX. This ends the proof.

Remark that our strategy can be used for many other types of contractions in partially ordered metric spaces under the continuity assumption of the considered mapping. For example, let us consider the Kannan fixed point theorem [3].

Theorem 2.1. Let (Z, d)be a complete metric space and T :Z → Z be a given mapping. Suppose that there exists some constantλ∈(0,1/2) such that

d(T x, T y)≤λ[d(x, T x) +d(y, T y)], (x, y)∈ Z × Z.

ThenT has a unique fixed.

Using our strategy, we deduce immediately from Theorem 2.1 the following version of Kannan fixed point theorem in partially ordered metric spaces.

Corollary 1. Let(X, d)be a complete metric space endowed with a partial order. LetT :X→X be a continuous and non-decreasing mapping with respect to . Suppose that there exists some constant λ∈(0,1/2) such that

d(T x, T y)≤λ[d(x, T x) +d(y, T y)], (x, y)∈X×X, xy.

If there exists somex0∈X such that x0T x0, thenT has a fixed point.

Competing interests The author declares that he has no competing interests.

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RAN-REURINGS FIXED POINT THEOREM 3

References

[1] S. Banach, Sur les op´erations dans les ensembles abstraits et leur applications aux ´equations int´egrales. Fundam.

Math. 3 (1922) 133–181.

[2] A.C.M. Ran, M.C.B. Reurings, A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc. 132 (2004) 1435–1443.

[3] R. Kannan, Some results on fixed points, Bull.Calcutta Math. Soc. 60 (1968) 71–76.

Bessem Samet

Department of Mathematics, College of Science

King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia E-mail: [email protected]

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