Fixed point results in ordered metric spaces for rational type expressions with auxiliary functions

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ORIGINAL ARTICLE

Fixed point results in ordered metric spaces for

rational type expressions with auxiliary functions

Sumit Chandok

a

, Binayak S. Choudhury

b

, Nikhilesh Metiya

c,

*

a

Department of Mathematics, Khalsa College of Engineering & Technology (Punjab Technical University), Ranjit Avenue, Amritsar 143001, India

b

Department of Mathematics, Bengal Engineering and Science University, Shibpur, Howrah 711103, West Bengal, India

c

Department of Mathematics, Bengal Institute of Technology, Kolkata 700150, West Bengal, India

Received 9 August 2013; revised 19 December 2013; accepted 2 February 2014 Available online 15 March 2014

KEYWORDS Fixed point;

Rational type generalized contraction mappings; Ordered metric spaces

Abstract In this paper, we establish some ﬁxed point results for mappings involvingð/; wÞ-rational

type contractions in the framework of metric spaces endowed with a partial order. These results generalize and extend some known results in the literature. Four illustrative examples are given.

2010 MATHEMATICS SUBJECT CLASSIFICATION: 41A50; 47H10; 54H25

ª 2014 Production and hosting by Elsevier B.V. on behalf of Egyptian Mathematical Society.

1. Introduction

The Banach’s contraction mapping principle is one of the most versatile elementary results of mathematical analysis. It is widely applied in different branches of mathematics and is re-garded as the source of metric ﬁxed point theory. There is a vast literature dealing with technical extensions and generaliza-tions of Banach’s contraction principle, some instances of these works are in[1–16].

In recent times, ﬁxed point theory has developed rapidly in partially ordered metric spaces, that is, metric spaces endowed

with a partial ordering. The theory originated at a relatively later point of time. An early result in this direction was estab-lished by Turinici in ordered metrizable uniform spaces[17]. Application of fixed point results in partially ordered metric spaces was made subsequently, for example, by Ran and Reurings[18]to solving matrix equations and by Nieto and Rodriguez-Lopez [19] to obtain solutions of certain partial differential equations with periodic boundary conditions. Recently, many researchers have obtained fixed point, common fixed point results in partially ordered metric spaces, some of which are in[20–33].

The purpose of this paper is to establish some ﬁxed point results satisfying a generalized contraction mapping of rational type in metric spaces endowed with partial order using some auxiliary functions. Four illustrative examples are given. 2. Mathematical preliminaries

In [34], Dass and Gupta proved the following ﬁxed point theorem.

* Corresponding author. Tel.: +91 9732559201.

E-mail addresses: [email protected], chandhok.sumit@gmail. com (S. Chandok), [email protected], [email protected]

(B.S. Choudhury),[email protected](N. Metiya). Peer review under responsibility of Egyptian Mathematical Society.

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Theorem 2.1 [34]. LetðX; dÞ be a complete metric space and T : X! X a mapping such that there exist a; b P 0 with aþ b < 1 satisfying

dðTx; TyÞ 6 a dðy; TyÞ ½1 þ dðx; TxÞ

1þ dðx; yÞ þ b dðx; yÞ;

for all x; y2 X: ð2:1Þ

Then T has a unique ﬁxed point.

In [35], Cabrera, Harjani and Sadarangani proved the above theorem in the context of partially ordered metric spaces.

Deﬁnition 2.1. SupposeðX; 6Þ is a partially ordered set and T : X! X. T is said to be monotone nondecreasing if for all x; y2 X,

x 6 y ) Tx 6 Ty: ð2:2Þ

Theorem 2.2 [35]. Let ðX; 6Þ be a partially ordered set and suppose that there exists a metric d in X such thatðX; dÞ is a complete metric space. Let T : X! X be a continuous and non-decreasing mapping such that(2.1)is satisﬁed for all x; y2 X with x 6 y. If there exists x02 X such that x06Tx0then T has

a ﬁxed point.

Theorem 2.3 [35]. Let ðX; 6Þ be a partially ordered set and suppose that there exists a metric d in X such thatðX; dÞ is a complete metric space. Assume that iffxng is a nondecreasing

sequence in X such that xn! x, then xn6x, for all n2 N.

Let T : X! X be a nondecreasing mapping such that(2.1)is satisﬁed for all x; y2 X with x 6 y. If there exists x02 X such

that x06Tx0then T has a ﬁxed point.

Theorem 2.4 [35]. In addition to the hypotheses of Theorem 2.2 (or Theorem2.3), suppose that for every x; y2 X, there exists u2 X such that u 6 x and u 6 y. Then T has a unique ﬁxed point.

Khan et al.[36]initiated the use of a control function that alters distance between two points in a metric space, which they called an altering distance function.

Deﬁnition 2.2 [36]. A function / :½0; 1Þ ! ½0; 1Þ is called an altering distance function if the following properties are satisﬁed:

(i) / is monotone increasing and continuous, (ii) /ðtÞ ¼ 0 if and only if t ¼ 0.

In our results in the following section we will use the follow-ing class of functions.

We denote

U¼ f/ : ½0; 1Þ ! ½0; 1Þ : / an altering distance functiong and W¼ fw : ½0; 1Þ ! ½0; 1Þ : for any sequence fxng in ½0; 1Þ

with xn! t > 0; lim wðxnÞ > 0g:

We note that W is nonempty. For, we deﬁne w on½0; 1Þ by wðtÞ ¼ et_{; t}_{2 ½0; 1Þ. Then w 2 W. Here we observe that}

wð0Þ ¼ 1 > 0. On the other hand, if wðtÞ ¼ t2_{; t}_{2 ½0; 1Þ, then}

w2 W and wð0Þ ¼ 0.

Note: For any w2 W, it is clear that wðtÞ > 0 for t > 0; and wð0Þ need not be equal to 0.

3. Main results

Theorem 3.1. LetðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Let T : X! X be a continuous and nondecreasing mapping such that for all x; y2 X with x 6 y,

/ðdðTx; TyÞÞ 6 /ðMðx; yÞÞ wðNðx; yÞÞ; ð3:1Þ

where/2 U; w 2 W,

Mðx; yÞ ¼ max dðy; TyÞ ½1 þ dðx; TxÞ 1þ dðx; yÞ ;

dðy; TxÞ ½1 þ dðx; TyÞ 1þ dðx; yÞ ; dðx; yÞ

and

Nðx; yÞ ¼ max dðy; TyÞ ½1 þ dðx; TxÞ

1þ dðx; yÞ ; dðx; yÞ

: If there exists x02 X with x06Tx0, then T has a ﬁxed point.

Proof. If Tx0¼ x0, then we have the result. Suppose that

x0< Tx0. Then we construct a sequencefxng in X such that

xnþ1¼ Txn; for every n P 0: ð3:2Þ

Since T is a nondecreasing mapping, we obtain by induc-tion that

x0< Tx0¼ x16Tx1¼ x26 6 Txn1¼ xn6Txn

¼ xnþ16 ð3:3Þ

If there exists n P 1 such that xnþ1¼ xn, then from (3.2), xnþ1¼ Txn¼ xn, that is, xn is a ﬁxed point of T and

the proof is ﬁnished. Suppose that xnþ1–xn, that is,

dðxnþ1; xnÞ–0, for all n P 1. Let Rn¼ dðxnþ1; xnÞ, for all

nP 0.

Since xn1< xn, for all n P 1, from(3.1), we have

/ðdðxn; xnþ1ÞÞ ¼ /ðdðTxn1; TxnÞÞ 6/ max dðxn; TxnÞ½1 þ dðxn1; Txn1Þ 1þ dðxn1; xnÞ ;dðxn; Txn1Þ½1 þ dðxn1; TxnÞ 1þ dðxn1; xnÞ ; dðxn1; xnÞ w max dðxn; TxnÞ½1 þ dðxn1; Txn1Þ 1þ dðxn1; xnÞ ; dðxn1; xnÞ ¼ / max dðxn; xnþ1Þ½1 þ dðxn1; xnÞ 1þ dðxn1; xnÞ ;dðxn; xnÞ½1 þ dðxn1; xnþ1Þ 1þ dðxn1; xnÞ ; dðxn1; xnÞ w max dðxn; xnþ1Þ½1 þ dðxn1; xnÞ 1þ dðxn1; xnÞ ; dðxn1; xnÞ ¼ /ðmaxfdðxn; xnþ1Þ; dðxn1; xnÞgÞ wðmaxfdðxn; xnþ1Þ; dðxn1; xnÞgÞ; that is, /ðRnÞ ¼ /ðmaxfRn; Rn1gÞ wðmaxfRn; Rn1gÞ: ð3:4Þ

If Rn> Rn1, then from(3.4), we have

/ðRnÞ 6 /ðRnÞ wðRnÞ; that is; wðRnÞ 6 0;

which is a contradiction. So, Rn6Rn1, that is, fRng is a

decreasing sequence. Then the inequality(3.4)yields that

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SincefRng is a decreasing sequence of positive real numbers

and is bounded below, there exists r P 0 such that

Rn¼ dðxnþ1; xnÞ ! r as n ! 1: ð3:6Þ

Now, we shall show that r¼ 0. Assume, to the contrary, that r > 0. Taking limit supremum in both sides of(3.5), using (3.6), the property of w and the continuity of /, we obtain

/ðrÞ 6 /ðrÞ þ lim ðwðRn1ÞÞ:

Since limðwðRn1ÞÞ ¼ lim wðRn1Þ, it follows that

/ðrÞ 6 /ðrÞ lim ðwðRn1ÞÞ; that is; lim ðwðRn1ÞÞ 6 0;

which, by the property of w, is a contradiction unless r¼ 0. Therefore,

Rn¼ dðxnþ1; xnÞ ! 0 as n ! 1: ð3:7Þ

Next we show thatfxng is a Cauchy sequence.

Suppose that fxng is not a Cauchy sequence. Then there

exists an > 0 for which we can ﬁnd two sequences of positive integersfmðkÞg and fnðkÞg such that for all positive integers k; nðkÞ > mðkÞ > k and dðxmðkÞ; xnðkÞÞ P . Assuming that

nðkÞ is the smallest such positive integer, we get

nðkÞ > mðkÞ > k; dðxmðkÞ; xnðkÞÞ P and dðxmðkÞ; xnðkÞ1Þ < :

Now,

6 dðxmðkÞ; xnðkÞÞ 6 dðxmðkÞ; xnðkÞ1Þ þ dðxnðkÞ1; xnðkÞÞ;

that is,

6 dðxmðkÞ; xnðkÞÞ 6 þ dðxnðkÞ1; xnðkÞÞ:

Letting k! 1 in the above inequality and using(3.7), we have lim k!1dðxmðkÞ; xnðkÞÞ ¼ : ð3:8Þ Again, dðxmðkÞ1; xnðkÞ1Þ 6 dðxmðkÞ1; xmðkÞÞ þ dðxmðkÞ; xnðkÞÞ þ dðxnðkÞ; xnðkÞ1Þ; and dðxmðkÞ; xnðkÞÞ 6 dðxmðkÞ; xmðkÞ1Þ þ dðxmðkÞ1; xnðkÞ1Þ þ dðxnðkÞ1; xnðkÞÞ:

Letting k! 1 in the above inequalities and using(3.7) and (3.8), we have lim k!1dðxmðkÞ1; xnðkÞ1Þ ¼ : ð3:9Þ Again, dðxnðkÞ1; xmðkÞÞ 6 dðxmðkÞ1; xnðkÞ1Þ þ dðxmðkÞ1; xmðkÞÞ and dðxmðkÞ1; xnðkÞ1Þ 6 dðxmðkÞ1; xmðkÞÞ þ dðxnðkÞ1; xmðkÞÞ:

Letting k! 1 in the above inequalities and using(3.7) and (3.9), we have lim k!1dðxnðkÞ1; xmðkÞÞ ¼ : ð3:10Þ Similarly, we have lim k!1dðxmðkÞ1; xnðkÞÞ ¼ : ð3:11Þ Let Mk¼ max dðxnðkÞ1; xnðkÞÞ½1 þ dðxmðkÞ1; xmðkÞÞ 1þ dðxmðkÞ1; xnðkÞ1Þ ; dðxnðkÞ1; xmðkÞÞ½1 þ dðxmðkÞ1; xnðkÞÞ 1þ dðxmðkÞ1; xnðkÞ1Þ ; dðxmðkÞ1; xnðkÞ1Þ ð3:12Þ and Nk¼ max dðxnðkÞ1; xnðkÞÞ½1 þ dðxmðkÞ1; xmðkÞÞ 1þ dðxmðkÞ1; xnðkÞ1Þ ; dðxmðkÞ1; xnðkÞ1Þ : ð3:13Þ Letting k! 1 in(3.12) and (3.13), using3.7, 3.9, 3.10 and 3.11, we have lim k!1Mk¼ maxf0; ; g ¼ ð3:14Þ and lim k!1Nk¼ maxf0; g ¼ : ð3:15Þ

Since xmðkÞ16xnðkÞ1, applying(3.1)and using(3.12) and (3.13), we have

/ðdðxmðkÞ; xnðkÞÞÞ ¼ /ðdðTxmðkÞ1; TxnðkÞ1ÞÞ

6_/ðM_k_{Þ wðN}_k_Þ:

Taking limit supremum in both sides of the above inequal-ity, using3.8, 3.14, 3.15, the property of w and the continuity of /, we obtain

/ðÞ 6 /ðÞ þ lim ðwðNkÞÞ:

Since limðwðNkÞÞ ¼ lim wðNkÞ, it follows that

/ðÞ 6 /ðÞ lim ðwðNkÞÞ; that is; lim ðwðNkÞÞ 6 0;

which, by(3.15)and the property of w, is a contradiction. Thus fxng is a Cauchy sequence in a complete metric space X.

Therefore, there exits u2 X such that limn!1 xn¼ u. Then

the continuity of T implies that Tu¼ Tðlimn!1 xnÞ ¼

lim_n!1 Txn¼ limn!1 xnþ1¼ u, that is, u is a ﬁxed point

of T. h

In our next theorem we relax the continuity assumption of the mapping T in Theorem 3.1 by imposing the following order condition of the metric space X:

Iffxng is a non-decreasing sequence in X such that xn! x,

then xn6x, for all n2 N.

Theorem 3.2. LetðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Assume that iffxng is a nondecreasing sequence in

X such that xn! x, then xn6x, for all n2 N. Let T : X ! X

be a nondecreasing mapping. Suppose that(3.1)holds, where Mðx; yÞ; Nðx; yÞ and the conditions upon ð/; wÞ are the same as in Theorem3.1. If there exists x02 X with x06Tx0, then T

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Proof. We take the same sequencefxng as in the proof of

The-orem 3.1. Then we have x06x16x26. . . 6 xn6xnþ16. . .,

that is,fxng is a nondecreasing sequence. Also, this sequence

converges to u. Then xn6u, for all n2 N.

Suppose that u–Tu, that is, dðu; TuÞ > 0. Let Mn¼ max dðu; TuÞ½1 þ dðxn; TxnÞ 1þ dðxn; uÞ ;dðu; TxnÞ½1 þ dðxn; TuÞ 1þ dðxn; uÞ ; dðxn; uÞ

¼ max dðu; TuÞ½1 þ dðxn; xnþ1Þ

1þ dðxn; uÞ ;dðu; xnþ1Þ½1 þ dðxn; TuÞ 1þ dðxn; uÞ ; dðxn; uÞ ð3:16Þ and Nn¼ max dðu; TuÞ½1 þ dðxn; TxnÞ 1þ dðxn; uÞ ; dðxn; uÞ

¼ max dðu; TuÞ½1 þ dðxn; xnþ1Þ 1þ dðxn; uÞ

; dðxn; uÞ

: ð3:17Þ

Letting n! 1 in (3.16) and (3.17), using the fact that xn! u as n ! 1, we have

lim

n!1Mn¼ maxfdðu; TuÞ; 0; 0g ¼ dðu; TuÞ > 0 ð3:18Þ

and lim

n!1Nn¼ maxfdðu; TuÞ; 0g ¼ dðu; TuÞ > 0: ð3:19Þ

Since xn6u for all n, applying (3.1) and using (3.16), (3.17), we have

/ðdðxnþ1; TuÞÞ ¼ /ðdðTxn; TuÞÞ 6 /ðMnÞ wðNnÞ:

Taking limit supremum in both sides of the above inequal-ity, using(3.18), (3.19), the property of w and the continuity of /, we obtain

/ðdðu; TuÞÞ 6 /ðdðu; TuÞÞ þ lim ðwðNnÞÞ:

Since limðwðNnÞÞ ¼ lim wðNnÞ, it follows that

/ðdðu; TuÞÞ 6 /ðdðu; TuÞÞlim ðwðNnÞÞ; that is; lim ðwðNnÞÞ 6 0;

which, by (3.19) and the property of w, is a contradiction. Hence, u¼ Tu, that is, u is a ﬁxed point of T. h

Now, we shall prove the uniqueness of the ﬁxed point. Theorem 3.3. In addition to the hypotheses of Theorem 3.1 (or Theorem 3.2), suppose that for every x; y2 X, there exists u2 X such that u 6 x and u 6 y. Then T has a unique ﬁxed point.

Proof. It follows from the Theorem 3.1 (or Theorem 3.2) that the set of ﬁxed points of T is non-empty. We shall show that if x _{and y} _{are two ﬁxed points of T, that is, if x}_{¼ Tx} _and

y_{¼ Ty}_{, then x}_{¼ y}_.

By the assumption, there exists u02 X such that u06x

and u06y. Then, similarly as in the proof of Theorem 3.1, we

deﬁne the sequencefung such that

unþ1¼ Tun¼ Tnþ1u0; n¼ 0; 1; 2; . . . ð3:20Þ

Monotonicity of T implies that

Tnu0¼ un6x¼ Tnxand Tnu0¼ un6y¼ Tny:

If there exists a positive integer m such that x_{¼ u} m, then

x_{¼ Tx}_{¼ Tu}

n¼ unþ1, for all n P m. Then un! x as

n! 1. Now we suppose that x_–u

n, for all n P 0. So

un< x, for all n P 0. Then dðun; xÞ–0, for all n P 0.

Let Pn¼ dðun; xÞ, for all n P 0. Since un< x, for all

nP 0, applying(3.1), we have /ðdðunþ1; xÞÞ ¼ /ðdðTun; TxÞÞ 6_{/ max} dðx _{; Tx}_{Þ½1 þ dðu} n; TunÞ 1þ dðun; xÞ ;dðx _{; Tu} nÞ½1 þ dðun; TxÞ 1þ dðun; xÞ ; dðun; xÞ w max dðx _{; Tx}_{Þ½1 þ dðu} n; TunÞ 1þ dðun; xÞ ; dðun; xÞ ¼ / max 0; dðx _{; u} nþ1Þ½1 þ dðun; xÞ 1þ dðun; xÞ ; dðun; xÞ wðmaxf0; dðun; xÞgÞ ¼ /ðmaxfdðx_{; u} nþ1Þ; dðun; xÞgÞ wðdðun; xÞÞ; that is, /ðPnþ1Þ ¼ /ðmaxfPnþ1; PngÞ wðPnÞ: ð3:21Þ

If Pnþ1> Pn, then from the above inequality, we have

/ðPnþ1Þ 6 /ðPnþ1Þ wðPnÞ; that is; wðPnÞ 6 0;

which is a contradiction. So, Pnþ16Pn, that is, fPng is a

decreasing sequence. Then it follows from the inequality (3.21)that

/ðPnþ1Þ 6 /ðPnÞ wðPnÞ: ð3:22Þ

SincefPng is a decreasing sequence of positive real numbers

and is bounded below, there exists r P 0 such that

Pn¼ dðun; xÞ ! r as n ! 1: ð3:23Þ

Arguing similarly as in the proof of Theorem 3.1 we can show that r¼ 0. Then,

Pn¼ dðun; xÞ ! 0 as n ! 1; that is; un! x as n

! 1: ð3:24Þ

Using a similar argument, we can prove that

un! y as n! 1: ð3:25Þ

Finally, the uniqueness of the limit implies x_{¼ y}_{. Hence}

Thas a unique ﬁxed point. h

Example 3.4. Let X¼ fð0; 1Þ; ð1; 0Þ; ð1; 1Þg R2

with the Euclidean distance d. We consider the partial order R in X as follows:

R¼ fðx; xÞ : x 2 Xg [ fðð0; 1Þ; ð1; 1ÞÞg: Let T : X! X be given by

Tð0; 1Þ ¼ ð0; 1Þ; Tð1; 0Þ ¼ ð1; 0Þ and Tð1; 1Þ ¼ ð0; 1Þ: Let /; w :½0; 1Þ ! ½0; 1Þ be given respectively by the formulas

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/ðtÞ ¼ t2 ; wðtÞ ¼ ½t2 2 ; if 3 < t < 4; t2 2; otherwise; (

where½t denotes the greatest integer not exceeding t. Here all the conditions of Theorems 3.1 and 3.2 are satisfied and T has two fixed points which areð0; 1Þ and ð1; 0Þ. Example 3.5. Let X¼ ð0; 0Þ; 1 2; 0 ; ð0; 1Þ be a subset of R2 _{with the order ‘‘6’’ defined as: for} _ðx

1; y1Þ; ðx2; y2Þ 2

X; ðx1; y1Þ 6 ðx2; y2Þ if and only if x16x2; y16y2. Let

d : X X ! R be given as

dðx; yÞ ¼ max fjx1 x2j; jy1 y2jg;

for x¼ ðx1; y1Þ; y ¼ ðx2; y2Þ 2 X:

Let T : X! X be deﬁned as follows:

Tð0; 0Þ ¼ ð0; 0Þ; Tð0; 1Þ ¼ ð1 2; 0Þ and T 1 2; 0 ¼ ð0; 0Þ: Let us consider the functions / and w as defined in Example 3.4. Here all the conditions of Theorems 3.1, 3.2 and 3.3 are satisfied andð0; 0Þ is the unique fixed point of T.

Example 3.6. Let X¼ ½1:5; 2 with usual partial order ‘‘6’’ and usual metric ‘‘d’’ be a partially ordered metric space. Let T : X! X be deﬁned as follows: Tx¼ 1:81; if 1:5 6 x < 1:75; xþ1 x 1 2; if 1:75 6 x 6 2:

Let /; w :½0; 1Þ ! ½0; 1Þ be given respectively by the formulas /ðtÞ ¼ t; wðtÞ ¼ ½t 1000; if 3 < t < 4; t 1000; otherwise; (

where½t denotes the greatest integer not exceeding t. Here all the conditions of Theorems 3.2 and 3.3 are satisﬁed and x¼ 2 is the unique ﬁxed point of T.

Note. In the above example the mapping T is not continu-ous. Therefore, the above example is not applicable to Theorem 3.1.

Corollary 3.7. LetðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Let T : X! X be a continuous and nondecreasing mapping such that for all x; y2 X with x 6 y,

/ðdðTx; TyÞÞ 6 /ðNðx; yÞÞ wðNðx; yÞÞ; ð3:26Þ

where Nðx; yÞ and the conditions upon ð/; wÞ are the same as in Theorem3.1. If there exists x02 X with x06Tx0, then T has a

ﬁxed point.

Proof. Since the inequality(3.26)implies the inequality(3.1), by Theorem 3.1, we have the required proof. h

Corollary 3.8. LetðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Assume that iffxng is a nondecreasing sequence in

X such that xn! x, then xn6x, for all n2 N. Let T : X ! X

be a nondecreasing mapping. Suppose that(3.26)holds, where Nðx; yÞ and the conditions upon ð/; wÞ are the same as in The-orem 3.1. If there exists x02 X with x06Tx0, then T has a

ﬁxed point.

Proof. Since the inequality(3.26)implies the inequality(3.1), by Theorem 3.2, we have the required proof. h

Corollary 3.9. In addition to the hypotheses of Corollary 3.7 (or Corollary 3.8), suppose that for every x; y2 X, there exists u2 X such that u 6 x and u 6 y. Then T has a unique ﬁxed point.

Example 3.10. Let X¼ C½0; 1 ¼ fx : ½0; 1 ! R; continuousg

with the partial order given by x 6 y() xðtÞ 6 yðtÞ,

for t2 ½0; 1. Let the metric d on X be given by dðx; yÞ ¼ Sup fjxðtÞ yðtÞj : t 2 ½0; 1g; for x; y 2 X:

Let T : X! X be deﬁned as Tx ¼x

3; x2 X.

Let /; w :½0; 1Þ ! ½0; 1Þ be given respectively by the formulas /ðtÞ ¼ t2_; _{wðtÞ ¼} ½t2 10 if 3 < t < 4; t2 10; otherwise; (

where½t denotes the greatest integer not exceeding t. Here, all the conditions of Corollaries 3.7 and 3.8 are satisfied. Moreover, as for x; y2 X ¼ C½0; 1, the function minðx; yÞðtÞ ¼ min fxðtÞ; yðtÞg is continuous, the conditions of Corollary 3.9 are satisfied. It is seen that x¼ 0 is the unique fixed point of T in X.

In the Corollaries 3.7 and 3.8, taking / to be the identity

mapping and wðtÞ ¼ ð1 kÞt for all t 2 ½0; 1Þ, where

k2 ð0; 1Þ, we have the following results.

Corollary 3.11. Let ðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Let T : X! X be a continuous and nondecreasing mapping. Suppose there exists k2 ð0; 1Þ such that for all x; y2 X with x 6 y,

dðTx; TyÞ 6 k max dðy; TyÞ½1 þ dðx; TxÞ

1þ dðx; yÞ ; dðx; yÞ

: ð3:27Þ If there exists x02 X with x06Tx0, then T has a ﬁxed point.

Corollary 3.12. LetðX; 6Þ be a partially ordered set and sup-pose that there exists a metric d on X such thatðX; dÞ is a com-plete metric space. Assume that if fxng is a nondecreasing

Let T : X! X be a nondecreasing mapping. Suppose that (3.27)holds. If there exists x02 X with x06Tx0, then T has

a ﬁxed point.

Remark 3.1. Theorems 3.1, 3.2 and 3.3 are respectively gener-alizations of Theorems 2, 3 and 4 in[35]which are also noted here as Theorems 2.2, 2.3 and 2.4 respectively.

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Other consequences of our results are the following for the mappings involving contractions of integral type.

Denote by K the set of functions l :½0; 1Þ ! ½0; 1Þ satisfy-ing the followsatisfy-ing hypotheses:

(h1) l is a Lebesgue-integrable mapping on each compact subset of½0; 1Þ;

(h2) for any > 0, we haveR₀lðtÞdt > 0.

Corollary 3.13. Let ðX; 6Þ be a partially ordered set and suppose that there exists a metric d on X such thatðX; dÞ is a complete metric space. Let T : X! X be a continuous and nondecreasing mapping. Suppose that there exists2 K and for all x; y2 X with x 6 y, Z /ðdðTx; TyÞÞ 0 sðtÞ dt 6 Z /ðMðx; yÞÞ 0 sðtÞ dt Z wðNðx; yÞÞ 0 sðtÞ dt; ð3:28Þ where Mðx; yÞ; Nðx; yÞ and the conditions upon ð/; wÞ are the same as in Theorem3.1. If there exists x02 X with x06Tx0,

then T has a ﬁxed point.

Let T : X! X be a nondecreasing mapping. Suppose that there exist s2 K for which (3.28) holds, where Mðx; yÞ; Nðx; yÞ and the conditions upon ð/; wÞ are the same as in Theorem 3.1. If there exists x02 X with x06Tx0, then T has a ﬁxed

point.

Corollary 3.15. LetðX; 6Þ be a partially ordered set and sup-pose that there exists a metric d on X such thatðX; dÞ is a

com-plete metric space. Let T : X! X be a continuous and

nondecreasing mapping. Suppose that there exists2 K and for all x; y2 X with x 6 y, Z /ðdðTx; TyÞÞ 0 sðtÞ dt 6 Z /ðNðx; yÞÞ 0 sðtÞ dt Z wðNðx; yÞÞ 0 sðtÞ dt; ð3:29Þ where Nðx; yÞ and the conditions upon ð/; wÞ are the same as in Theorem3.1. If there exists x02 X with x06Tx0, then T has a

ﬁxed point.

Let T : X! X be a nondecreasing mapping. Suppose that there exists2 K for which(3.29)holds, where Nðx; yÞ and the con-ditions uponð/; wÞ are the same as in Theorem 3.1. If there exists x02 X with x06Tx0, then T has a ﬁxed point.

Acknowledgment

The authors gratefully acknowledge the suggestions made by the learned referee.

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