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Research Article

Various Suzuki type theorems in b-metric spaces

A. Latifa,∗, V. Parvanehb, P. Salimic, A. E. Al-Mazrooeid

aDepartment of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia.

bDepartment of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, Gilan-E-Gharb, Iran.

cYoung Researchers and Elite Club, Rasht Branch, Islamic Azad University, Rasht, Iran.

dDepartment of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia.

Communicated by N. Hussain

Abstract

In this paper, we prove some fixed point results for α-admissible mappings which satisfy Suzuki type con- tractive condition in the setup of b-metric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results. c2015 All rights reserved.

Keywords: Ordered metric space, b-metric space, fixed point.

2010 MSC: 47H10, 54H25.

1. Introduction

Banach contractive principle or Banach fixed point theorem is the most celebrated result in fixed point theory which illustrates that in a complete metric space, each contractive mapping has a unique fixed point. There is a great number of generalizations of Banach contraction principle by using different forms of contractive conditions in various spaces. Some of such generalizations are obtained by contraction conditions described by rational expressions, (see, [14, 18, 20, 25]).

Ran and Reurings initiated the studying of fixed point results on partially ordered sets in [21], where they gave many useful results in matrix equations. Recently, many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order or a graph and obtained many fixed point results in such spaces. For more details on fixed point results, their applications, comparison of different contractive conditions and related results in ordered metric spaces and spaces endowed with a graph we refer the reader to [5] and [19].

Corresponding author

Email addresses: [email protected](A. Latif),[email protected](V. Parvaneh),[email protected] (P. Salimi),[email protected](A. E. Al-Mazrooei)

Received 2014-09-29

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Czerwik in [8] introduced the concept of a b-metric space. Since then, several papers dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see, [7, 9, 16, 22]).

Definition 1.1. LetX be a (nonempty) set ands≥1 be a given real number. A functiond:X×X →R+ is ab-metric if, for all x, y, z∈X, the following conditions are satisfied:

(b1) d(x, y) = 0 iffx=y, (b2) d(x, y) =d(y, x),

(b3) d(x, z)≤s[d(x, y) +d(y, z)].

In this case, the pair (X, d) is called a b-metric space.

Definition 1.2 ([6]). Let (X, d) be a b-metric space.

(a) A sequence {xn}inX is calledb-convergent if and only if there existsx∈X such thatd(xn, x)→0, asn→ ∞. In this case, we write lim

n→∞xn=x.

(b){xn} inX is said to be b-Cauchy if and only if d(xn, xm)→0,asn, m→ ∞.

(c) Theb-metric space (X, d) isb-complete if every b-Cauchy sequence inX isb-converges.

Note that ab-metric need not to be a continuous function. The following example (corrected from [13]) illustrates this fact.

Example 1.3. Let X=N∪ {∞}and let d:X×X→Rbe defined by

d(m, n) =









0, ifm=n,

|m11n|, if one of m, n is even and the other is even or∞,

5, if one of m, n is odd and the other is odd (andm6=n) or ∞, 2, otherwise.

It can be checked that for allm, n, p∈X, we have d(m, p)≤ 5

2(d(m, n) +d(n, p)).

Thus, (X, d) is a b-metric space (with s= 5/2). Let xn= 2nfor each n∈N. Then d(2n,∞) = 1

2n →0 as n→ ∞, that is,xn→ ∞, butd(xn,1) = 26→5 =d(∞,1) asn→ ∞.

Lemma 1.4([1]). Let(X, d)be ab-metric space withs≥1, and suppose that{xn}and{yn}areb-convergent tox andy, respectively. Then we have

1

s2d(x, y)≤lim inf

n→∞ d(xn, yn)≤lim sup

n→∞ d(xn, yn)≤s2d(x, y).

In particular, if x=y, then we have lim

n→∞d(xn, yn) = 0. Moreover, for each z∈X, we have, 1

sd(x, z)≤lim inf

n→∞ d(xn, z)≤lim sup

n→∞ d(xn, z)≤sd(x, z).

Let Sdenotes the class of all real functionsβ : [0,+∞)→[0,1) satisfying the condition β(tn)→1 implies that tn→0, as n→ ∞.

In order to generalize the Banach contraction principle, in 1973, Geraghty proved the following.

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Theorem 1.5 ([12]). Let (X, d) be a complete metric space, and letf :X →X be a self-map. Suppose that there existsβ ∈S such that

d(f x, f y)≤β(d(x, y))d(x, y)

holds for allx, y∈X. Then f has a unique fixed pointz∈X and for each x∈X the Picard sequence{fnx}

converges to z.

In [10], some fixed point theorems for Geraghty-type contractive mappings in various generalized metric spaces are proved. As in [10], we will consider the class of functions F, whereβ ∈ F ifβ: [0,∞)→[0,1/s) has the property

β(tn)→ 1

s implies that tn→0, as n→ ∞.

Theorem 1.6 ([10]). Let s >1, and let (X, D, s) be a complete metric type space. Suppose that a mapping f :X →X satisfies the condition

D(f x, f y)≤β(D(x, y))D(x, y)

for all x, y ∈ X and some β ∈ F. Then f has a unique fixed point z ∈X, and for each x ∈ X the Picard sequence{fnx} converges to z in (X, D, s).

Unification of the recent results of Zabihi and Razani [28] yield the following result.

Theorem 1.7. Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that(X, d) is ab-complete b-metric space (with parameter s >1). Letf :X→X be an increasing mapping with respect tosuch that there exists an elementx0∈X withx0 f(x0). Suppose that there exists β∈ F such that,

sd(f x, f y)≤β(d(x, y))M(x, y) +LN(x, y) (1.1) for all comparable elements x, y∈X, where L≥0,

M(x, y) = max

d(x, y),d(x, f x)d(y, f y) 1 +d(f x, f y)

and

N(x, y) = min{d(x, f x), d(x, f y), d(y, f x), d(y, f y)}.

If f is continuous, or, whenever {xn} is a nondecreasing sequence in X such that xn → u ∈ X, one has xn u for all n∈N, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only iff has one and only one fixed point.

One of the interesting results which generalizes the Banach contraction principle was given by Sametet al. [23] by definingα-ψ-contractive mappings.

Definition 1.8 ([23]). Let T be a self-mapping on X and letα :X×X → [0,∞) be a function. We say thatT is anα-admissible mapping if

x, y∈X, α(x, y)≥1 =⇒ α(T x, T y)≥1.

Denote with Ψ0 the family of all nondecreasing functionsψ: [0,∞)→[0,∞) such thatP

n=1ψn(t)<∞ for all t >0, where ψn is the n-th iterate of ψ.

Theorem 1.9 ([23]). Let (X, d) be a complete metric space and let T be an α-admissible mapping. Assume that

α(x, y)d(T x, T y)≤ψ(d(x, y)) (1.2)

where ψ∈Ψ0. Also, suppose that the following assertions hold:

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(i) there exists x0∈X such that α(x0, T x0)≥1;

(ii) either T is continuous, or, for any sequence {xn} in X withα(xn, xn+1)≥1 for all n∈N∪ {0} such thatxn→x as n→ ∞, we have α(xn, x)≥1 for all n∈N∪ {0}.

Then T has a fixed point.

Definition 1.10 ([17]). Let f : X → X and α : X ×X → [0,+∞). We say that f is a triangular α-admissible mapping if

(T1) α(x, y)≥1 implies α(f x, f y)≥1, x, y∈X,

(T2)

α(x, z)≥1

α(z, y)≥1 implies α(x, y)≥1, x, y, z ∈X.

Example 1.11 ([17]). Let X = R, f x = √3

x and α(x, y) = ex−y, then f is a triangular α-admissible mapping. Indeed, if α(x, y) = ex−y ≥ 1, then x ≥ y which implies that f x ≥ f y, that is, α(f x, f y) = ef x−f y≥1. Also, if

α(x, z)≥1

α(z, y)≥1 , then

x−z≥0,

z−y≥0, that is, x−y≥0 and so,α(x, y) =ex−y ≥1.

Lemma 1.12 ([17]). Let f be a triangular α-admissible mapping. Assume that there exists x0 ∈ X such thatα(x0, f x0)≥1. Define sequence {xn} by xn=fnx0. Then

α(xm, xn)≥1 for all m, n∈N withm < n.

We now recall the concept of (c)-comparison function which was introduced by Berinde [4].

Definition 1.13 ([4]). A function ϕ: [0,∞)→[0,∞) is said to be a (c)-comparison function if (c1) ϕis increasing,

(c2) there existsk0 ∈N,a∈(0,1) and a convergent series of nonnegative terms

X

k=1

vksuch that ϕk+1(t)≤ aϕk(t) +vk, fork≥k0 and any t∈[0,∞).

Later, Berinde [3] introduced the notion of a (b)-comparison function as a generalization of the concept of (c)-comparison function.

Definition 1.14([3]). Lets≥1 be a real number. A mappingϕ: [0,∞)→[0,∞) is called a (b)-comparison function if the following conditions are fulfilled

(1) ϕis monotone increasing;

(2) there existk0 ∈N,a∈(0,1) and a convergent series of nonnegative terms

X

k=1

vksuch thatsk+1ϕk+1(t)≤ askϕk(t) +vk for any k≥k0 and anyt∈[0,∞).

Let Ψb be the class of all (b)-comparison functions ϕ : [0,∞) → [0,∞). It is clear that the notion of (b)-comparison function coincide with (c)-comparison function fors= 1.

We now recall the following lemma which will simplify the proofs.

Lemma 1.15 ([2]). If ϕ: [0,∞)→[0,∞) is a (b)-comparison function, then we have the following.

(1) the series

X

k=0

skϕk(t) converges for anyt∈R+;

(2) the functionbs: [0,∞)→[0,∞)defined bybs(t) =

X

k=0

skϕk(t), t∈[0,∞), is increasing and continuous at0.

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2. Main Results

In 1962, Edelstein [11] proved an interesting version of Banach contraction principle. In 2009, Suzuki [27] proved certain remarkable results to improve the results of Banach and Edelstein (see also [24, 26]).

Now, we are ready to prove the following Suzuki type theorems for nonlinear contractions.

Theorem 2.1. Let (X, d) be a b-complete b-metric space (with parameter s >1) and let f be a triangular α-admissible mapping. Suppose that there exists β ∈ F such that,

1

2sd(x, f x)≤d(x, y) =⇒sα(x, y)d(f x, f y)≤β(M(x, y))M(x, y) (2.1) for allx, y∈X, where

M(x, y) = max

d(x, y),d(x, f x)d(x, f y) +d(y, f y)d(y, f x) 1 +s[d(x, y) +d(f x, f y)] , d(x, f x)d(x, f y) +d(y, f y)d(y, f x)

1 +d(x, f y) +d(y, f x)

. Also, suppose that the following assertions hold:

(i) there exists x0∈X such that α(x0, f x0)≥1;

(ii) for any sequence {xn} in X withα(xn, xn+1)≥1for all n∈N∪ {0}such that xn→x asn→ ∞, we haveα(xn, x)≥1 for alln∈N∪ {0}.

Then,T has a fixed point.

Proof. Let x0 ∈X be such thatα(x0, f x0)≥1. Define a sequence{xn} by xn=fnx0 for all n∈N. Since f is anα-admissible mapping and α(x0, x1) =α(x0, f x0)≥1, we deduce that α(x1, x2) =α(f x0, f x1)≥1.

Continuing this process, we get that α(xn, xn+1) ≥ 1 for all n ∈ N∪ {0}. We will do the proof in the following steps.

Step I:We will show that lim

n→∞d(xn, xn+1) = 0. Sinceα(xn, xn+1)≥1 for eachn∈N, and2s1d(xn−1, f xn−1)≤ d(xn−1, xn) then by (2.1) we have

d(xn, xn+1) =d(f xn−1, f xn)

≤sα(xn−1, xn)d(f xn−1, f xn)

≤β(M(xn−1, xn))M(xn−1, xn)

≤β(d(xn−1, xn))d(xn−1, xn)

< 1sd(xn−1, xn)

≤d(xn−1, xn),

(2.2)

because

M(xn−1, xn) = max{d(xn−1, xn),

d(xn−1, f xn−1)d(xn−1, f xn) +d(xn, f xn)d(xn, f xn−1) 1 +s[d(xn−1, xn) +d(f xn−1, f xn)] , d(xn−1, f xn−1)d(xn−1, f xn) +d(xn, f xn)d(xn, f xn−1)

1 +d(xn−1, f xn) +d(xn, f xn−1) }

= max{d(xn−1, xn),

d(xn−1, xn)d(xn−1, xn+1) +d(xn, xn+1)d(xn, xn) 1 +s[d(xn−1, xn) +d(xn, xn+1)] , d(xn−1, xn)d(xn−1, xn+1) +d(xn, xn+1)d(xn, xn)

1 +d(xn−1, xn+1) +d(xn, xn) }

=d(xn−1, xn).

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Therefore, {d(xn, xn+1)} is decreasing. Then there exists r ≥ 0 such that lim

n→∞d(xn, xn+1) = r. We will prove thatr = 0. Suppose on contrary thatr >0. Then, lettingn→ ∞, from (2.2) we have

1

sr≤ lim

n→∞β(d(xn−1, xn))r which implies that d(xn−1, xn)→0. Hence, r= 0, a contradiction. So,

n→∞lim d(xn−1, xn) = 0 (2.3)

holds true.

Step II: Now, we prove that the sequence{xn}is ab-Cauchy sequence. Suppose the contrary, i.e., that {xn} is not a b-Cauchy sequence. Then there exists ε > 0 for which we can find two subsequences {xmi} and {xni} of {xn}such that ni is the smallest index for which

ni > mi> i andd(xmi, xni)≥ε. (2.4) This means that

d(xmi, xni−1)< ε. (2.5)

From ((2.4)) and using the triangular inequality, we get

ε≤d(xmi, xni)≤sd(xmi, xmi+1) +sd(xmi+1, xni).

Taking the upper limit as i→ ∞, we get ε

s ≤lim sup

i→∞

d(xmi+1, xni). (2.6)

Remember that, from (2.2) we get,

d(xn, xn+1)< d(xn−1, xn) (2.7)

for all n∈N. Suppose that there exists i0 ∈N such that, 1

2sd(xmi

0, f xmi

0)> d(xmi

0, xni

0−1)

and 1

2sd(xmi

0+1, f xmi

0+1)> d(xmi

0+1, xni

0−1).

Then from (2.7) we have,

d(xmi0, xmi0+1)≤s[d(xmi0, xni0−1) +d(xmi0+1, xni0−1)]

< s[ 1 2sd(xmi

0, f xmi

0) + 1 2sd(xmi

0+1, f xmi

0+1)]

= 1 2[d(xmi

0, xmi

0+1) +d(xmi

0+1, xmi

0+2)]

≤ 1 2[d(xmi

0, xmi

0+1) +d(xmi

0, xmi

0+1)] =d(xmi

0, xmi

0+1) which is a contradiction. Hence, either,

1

2sd(xmi, f xmi)≤d(xmi, xni−1)

or 1

2sd(xmi+1, f xmi+1)≤d(xmi+1, xni−1)

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holds for alli∈N. First suppose that

1

2sd(xmi, f xmi)≤d(xmi, xni−1).

As from Lemma 1.12,α(xmi, xni−1)≥1, we obtain that s·ε

s ≤s·lim sup

i→∞

d(xmi+1, xni)≤lim sup

i→∞

β(M(xmi, xni−1)) lim sup

i→∞

M(xmi, xni−1)

≤εlim sup

i→∞

β(M(xmi, xni−1)) because, from the definition ofM(x, y) and the above limits,

lim sup

i→∞

M(xmi, xni−1) = lim sup

i→∞

max{d(xmi, xni−1),

d(xmi, f xmi)d(xmi, f xni−1) +d(xni−1, f xni−1)d(xni−1, f xmi) 1 +s[d(xmi, xni−1) +d(f xmi, f xni−1)] , d(xmi, f xmi)d(xmi, f xni−1) +d(xni−1, f xni−1)d(xni−1, f xmi)

1 +d(xmi, f xni−1) +d(xni−1, f xmi) }

= lim sup

i→∞

max{d(xmi, xni−1),

d(xmi, xmi+1)d(xmi, xni) +d(xni−1, xni)d(xni−1, xmi+1) 1 +s[d(xmi, xni−1) +d(xmi+1, xni)] , d(xmi, xmi+1)d(xmi, xni) +d(xni−1, xni)d(xni−1, xmi+1)

1 +d(xmi, xni) +d(xni−1, xmi+1) }

≤ε, which implies that 1s ≤ lim sup

i→∞

β(M(xmi, xni−1)). Now, as β ∈ F we conclude that M(xmi, xni−1) → 0, hence, we get thatd(xmi, xni−1) → 0 which implies that d(xmi, xni)→ 0, a contradiction to 2.4. So, {xn} is a b-Cauchy sequence. b-Completeness ofX yields that{xn} b-converges to a pointx ∈X.

On the other hand, from ((2.4)) and using the triangular inequality, we get ε

s ≤d(xmi, xni)≤sd(xmi, xmi+2) +sd(xmi+2, xni).

Taking the upper limit as i→ ∞, we get ε

s ≤lim sup

i→∞

d(xmi+2, xni). (2.8)

Also, from ((2.6)) and using the triangular inequality, we get

d(xmi+1, xni−1)≤sd(xmi+1, xni) +sd(xni, xni−1).

Taking the upper limit as i→ ∞, we get lim sup

i→∞

d(xmi+1, xni−1)≤sε. (2.9)

Now, let

1

2sd(xmi+1, f xmi+1)≤d(xmi+1, xni−1).

From Lemma 1.12,α(xmi+1, xni−1)≥1, so, we have s·ε

s ≤s·lim sup

i→∞

d(xmi+2, xni)≤lim sup

i→∞

β(M(xmi+1, xni−1)) lim sup

i→∞

M(xmi+1, xni−1)

≤sεlim sup

i→∞

β(M(xmi+1, xni−1))

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because, lim sup

i→∞

M(xmi+1, xni−1) = lim sup

i→∞

max{d(xmi+1, xni−1),

d(xmi+1, f xmi+1)d(xmi+1, f xni−1) +d(xni−1, f xni−1)d(xni−1, f xmi+1) 1 +s[d(xmi+1, xni−1) +d(f xmi+1, f xni−1)] , d(xmi+1, f xmi+1)d(xmi+1, f xni−1) +d(xni−1, f xni−1)d(xni−1, f xmi+1)

1 +d(xmi+1, f xni−1) +d(xni−1, f xmi+1) }

= lim sup

i→∞

max{d(xmi+1, xni−1),

d(xmi+1, xmi+2)d(xmi+1, xni) +d(xni−1, xni)d(xni−1, xmi+2) 1 +s[d(xmi+1, xni−1) +d(xmi+2, xni)] , d(xmi+1, xmi+2)d(xmi+1, xni) +d(xni−1, xni)d(xni−1, xmi+2)

1 +d(xmi+1, xni) +d(xni−1, xmi+2) }

≤sε, which implies that 1s ≤lim sup

i→∞

β(M(xmi+1, xni−1)). Now, as β∈ F we conclude thatM(xmi+1, xni−1)→0, hence, we get thatd(xmi+1, xni−1)→ 0 which implies that d(xmi, xni) →0, a contradiction. So, {xn} is a b-Cauchy sequence. b-Completeness ofX yields that{xn} b-converges to a pointx ∈X.

Remember that, from (2.2) we get,

d(xn, xn+1)< d(xn−1, xn) (2.10)

for all n∈N. Suppose that there exists n0∈Nsuch that, 1

2sd(xn0, f xn0)> d(xn0, x)

and 1

2sd(xn0+1, f xn0+1)> d(xn0+1, x).

Then, from (2.10) we have,

d(xn0, xn0+1)≤s[d(xn0, x) +d(xn0+1, x)]

< s[1

2sd(xn0, f xn0) + 1

2sd(xn0+1, f xn0+1)]

= 1

2[d(xn0, xn0+1) +d(xn0+1, xn0+2)]

≤ 1

2[d(xn0, xn0+1) +d(xn0, xn0+1)] =d(xn0, xn0+1) which is a contradiction. Hence, either,

1

2sd(xn, f xn)≤d(xn, x)

or 1

2sd(xn+1, f xn+1)≤d(xn+1, x) holds for alln∈N. First, suppose that,

1

2sd(xn, f xn)≤d(xn, x)

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holds for alln∈N. Then from (2.1) we have,

d(f x, x)≤s[d(f x, f xn) +d(f xn, x)]

=s[d(f x, f xn) +d(xn+1, x)]

≤s[β(M(x, xn))M(x, xn) +d(xn+1, x)]

for all n∈N. Taking the limit as n→ ∞ in the above inequality we get,d(f x, x) = 0.i.e., f x =x.

By a similar method we can deducef x =x when 1

2d(xn+1, f xn+1)≤d(xn+1, x).

Hence, we proved thatx is a fixed point off.

Theorem 2.2. Let (X, d) be a b-complete b-metric space and let f be an α-admissible mapping. Suppose that there exists ψ∈Ψb such that,

1

2sd(x, f x)≤d(x, y) =⇒α(x, y)d(f x, f y)≤ψ(M(x, y)) (2.11) where,

M(x, y) = max

d(x, y), d(x, f x)d(y, f y)

1 +s[d(x, y) +d(x, f y)], d(x, f y)d(x, y) 1 +s[d(x, f x) +d(y, f y)]

, for allx, y∈X.

Also, suppose that the following assertions hold:

(i) there exists x0∈X such that α(x0, f x0)≥1;

(ii) for any sequence {xn} in X withα(xn, xn+1)≥1for all n∈N∪ {0}such that xn→x asn→ ∞, we haveα(xn, x)≥1 for alln∈N∪ {0}.

Then,f has a fixed point.

Proof. Let x0 ∈X be such thatα(x0, f x0)≥1. Define a sequence{xn} by xn=fnx0 for all n∈N. Since f is anα-admissible mapping and α(x0, x1) =α(x0, f x0)≥1, we deduce that α(x1, x2) =α(f x0, f x1)≥1.

Continuing this process, we get that α(xn, xn+1)≥1 for alln∈N∪ {0}.

If there existsn0 ∈Nsuch thatxn0 =xn0+1 then,xn0 =f xn0 and so we have no thing for prove. Hence, for all n∈N we assume thatd(xn, xn+1)>0.

On the other hand, we have, 1

2sd(xn−1, f xn−1) = 1

2d(xn−1, xn)≤d(xn−1, xn) and α(xn, xn+1)≥1, so by (2.11) we get,

d(xn, xn+1) =d(f xn−1, f xn)≤α(xn−1, xn)d(f xn−1, f xn)≤ψ(M(xn−1, xn)) (2.12) where,

M(xn−1, xn) = max{d(xn−1, xn), d(xn−1, f xn−1)d(xn, f xn) 1 +s[d(xn−1, xn) +d(xn−1, f xn)], d(xn−1, f xn)d(xn−1, xn)

1 +s[d(xn−1, f xn−1) +d(xn, f xn)]}

= max{d(xn−1, xn), d(xn−1, xn)d(xn, xn+1) 1 +s[d(xn−1, xn) +d(xn−1, xn+1)], d(xn−1, xn+1)d(xn−1, xn)

1 +s[d(xn−1, xn) +d(xn, xn+1)]}

=d(xn−1, xn).

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Hence,

d(xn, xn+1)≤ψ(d(xn−1, xn))< d(xn−1, xn). (2.13) By induction, we get that

d(xn, xn+1)≤ψ(d(xn−1, xn))≤ψ2(d(xn−2, xn−1))≤ · · · ≤ψn(d(x0, x1)). (2.14) Then, by the triangular inequality and (2.14), we get

d(xn, xm)≤sd(xn, xn+1) +s2d(xn+1, xn+2) +· · ·+sm−n−1d(xm−1, xm)

m−2

X

k=n

sk−n+1ψk(d(x0, x1))

X

k=n

skψk(d(x0, x1))−→0, asn−→ ∞.

Hence,{xn}is a b-Cauchy sequence. b-completeness ofX yields that{xn}converges to a pointx∈X, that is,xn→x asn→ ∞.

On the other hand, from (2.13) we get,

d(xn, xn+1)< d(xn−1, xn) (2.15)

for all n∈N. Suppose that there exists n0∈Nsuch that, 1

2sd(xn0, f xn0)> d(xn0, x)

and 1

2sd(xn0+1, f xn0+1)> d(xn0+1, x).

Then from (2.15) we have,

d(xn0, xn0+1)≤s[d(xn0, x) +d(xn0+1, x)]

< s[1

2sd(xn0, f xn0) + 1

2sd(xn0+1, f xn0+1)]

= 1

2[d(xn0, xn0+1) +d(xn0+1, xn0+2)]

≤ 1

2[d(xn0, xn0+1) +d(xn0, xn0+1)] =d(xn0, xn0+1) which is a contradiction. Hence, either,

1

2sd(xn, f xn)≤d(xn, x)

or 1

2sd(xn+1, f xn+1)≤d(xn+1, x) holds for alln∈N. First, suppose that,

1

2sd(xn, f xn)≤d(xn, x) holds for alln∈N. Then from (2.11) and hypothesis (ii), we have,

d(T x, x) ≤ s[d(f x, f xn) +d(f xn, x)]

≤ s[α(x, xn)d(f x, f xn) +d(xn+1, x)]

≤ s[ψ(M(x, xn)) +d(xn+1, x)]

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for all n∈N. Taking the limit as n→ ∞ in the above inequality we get,d(f x, x) = 0.i.e., f x =x.

By a similar method we can deducef x =x when 1

2d(xn+1, f xn+1)≤d(xn+1, x).

Hence, we proved thatx is a fixed point off.

Example 2.3. Let X=R2. We defineα:X×X →[0,∞) by

α(x, y) =

1, x, y∈U ={(0,0),(4,0),(0,4),(4,5),(5,4)}

0, otherwise.

Define metricdonXbyd((x1, x2),(y1, y2)) = (x1−y1)2+ (x2−y2)2. Clearly, (X, d,2) is a completeb-metric space. Also, define f :X→X andψ: [0,∞)→[0,∞) by

f(x1, x2) =









(x1,0), If x1 ≤x2 and x1, x2∈U (0, x2) If x1 > x2 and x1, x2∈U (2x21,3x32) If x1, x2∈R2\U

and ψ(t) = 0.99t.

First we assume that 14d(x, f x)≤d(x, y) andα(x, y)≥1. Then, (x, y)∈ n

(0,0),(4,0)

,

(0,0),(0,4)

,

(0,0),(4,5)

,

(0,0),(5,4)

,

(4,0),(0,0)

,

(4,0),(0,4)

,

(4,0),(5,4)

,

(4,0),(4,5)

,

(0,4),(0,0)

,

(0,4),(4,0)

,

(0,4),(5,4)

,

(0,4),(4,5)

,

(4,5),(0,0)

,

(4,5),(4,0)

,

(4,5),(0,4)

,

(4,5),(5,4)

(5,4),(0,0)

,

(5,4),(4,0)

,

(5,4),(0,4)

,

(5,4),(4,5) o

.

Since, d(f x, f y) =d(f y, f x) and d(x, y) =d(y, x), hence without any loss of generality we can reduce the above set to the following:

(x, y)∈ n

(0,0),(4,0)

,

(0,0),(0,4)

,

(0,0),(4,5)

,

(0,0),(5,4)

,

(4,0),(0,4)

,

(4,0),(5,4)

,

(4,0),(4,5)

,

(0,4),(5,4)

,

(0,4),(4,5) o

. Now, we consider the following cases:

• Let (x, y) =

(0,0),(4,0)

, then,

d(f x, f y) =d(f(0,0), f(4,0)) = 0≤ψ(d(x, y)).

• Let (x, y) =

(0,0),(0,4)

, then,

d(f x, f y) =d(f(0,0), f(0,4)) = 0≤ψ(d(x, y)).

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• Let (x, y) =

(0,0),(4,5)

, then,

d(f x, f y) =d(f(0,0), f(4,5)) = 16≤40.59 = 0.99×41 =ψ(d((0,0),(4,5))).

• Let (x, y) =

(0,0),(5,4)

, then,

d(f x, f y) =d(f(0,0), f(5,4)) = 16≤40.59 = 0.99×41 =ψ(d((0,0),(4,5))).

• Let (x, y) =

(4,0),(0,4)

, then,

d(f x, f y) =d(f(4,0), f(0,4)) = 0≤ψ(d(x, y)).

• Let (x, y) =

(4,0),(5,4)

, then,

d(f x, f y) =d(f(4,0), f(5,4)) = 16≤16.83 = 0.99×17 =ψ(d((4,0),(5,4))).

• Let (x, y) =

(4,0),(4,5)

, then,

d(f x, f y) =d(f(4,0), f(4,5)) = 16≤16.83 = 0.99×17 =ψ(d((4,0),(4,5))).

• Let (x, y) =

(0,4),(5,4)

, then,

d(f x, f y) =d(f(0,4), f(5,4)) = 16≤24.75 = 0.99×25 =ψ(d((0,4),(5,4))).

• Let (x, y) =

(0,4),(4,5)

, then,

d(f x, f y) =d(f(0,4), f(4,5)) = 16≤16.83 = 0.99×17 =ψ(d((0,4),(4,5))).

That is, 14d(x, f x) ≤ d(x, y) and α(x, y) ≥ 1 implies that d(f x, f y) ≤ ψ(d(x, y)). Let α(x, y) ≥ 1, then x, y ∈U. On the other hand, f w ∈ U for all w ∈ U. Then, α(f x, f y) ≥1. That is, f is an α-admissible mapping. If {xn} be a sequence inX such that α(xn, xn+1)≥1 withxn →x asn→ ∞, then,xn∈U for all n∈ N. Also, U is a closed set. Then, x∈ [0,+∞). That is, α(xn, x) ≥1 for all n∈ N∪ {0}. Clearly, α((0,0), f(0,0))≥1.

Therefore all conditions of Theorem 2.2 holds and f has a fixed point. Here,x = (0,0) is a fixed point off.

3. Contractive mappings on b-metric spaces endowed with a graph

Recently, some results have appeared for a mapping to be a Picard Operator where (X, d) is endowed with a graph. The first result in this direction was given by Jachymski [15].

Definition 3.1 ([15]). Let (X, d) be a metric space endowed with a graph G. We say that a self-mapping f :X →X is a BanachG-contraction or simply a G-contraction if f preserves the edges of G, that is,

(x, y)∈E(G) =⇒(f x, f y)∈E(G) for all x, y∈X and f decreases the weights of the edges of Gin the following way:

∃α∈(0,1) such that for allx, y∈X, (x, y)∈E(G) =⇒d(f x, f y)≤αd(x, y).

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Definition 3.2. Let (X, d) be a b-metric space endowed with a graph G. We say that a self-mapping f :X →X is aG-ψ-Suzuki type rational contraction if T preserves the edges of G, that is,

(x, y)∈E(G) =⇒(f x, f y)∈E(G), for all x, y∈X and f decreases the weights of the edges of Gin the following way:

1

2sd(x, f x)≤d(x, y) =⇒d(f x, f y)≤ψ(M(x, y)) (3.1) where,

M(x, y) = max

d(x, y), d(x, f x)d(y, f y)

1 +s[d(x, y) +d(x, f y)], d(x, f y)d(x, y) 1 +s[d(x, f x) +d(y, f y)]

and ψ∈Ψb for all (x, y)∈E(G).

Theorem 3.3. Let (X, d)be a b-completeb-metric space endowed with a graphG. Assume that T :X→X is a G-ψ-Suzuki type rational contraction such that the following conditions hold:

(i) there exists an element x0∈X such that (x0, f x0)∈E(G);

(ii) for any sequence {xn} in X with(xn, xn+1)∈E(G) for alln∈N∪ {0} such that xn→x as n→ ∞, we have (xn, x)∈E(G) for all n∈N∪ {0}.

ThenT has a fixed point.

Proof. Define α:X×X→[0,+∞) by α(x, y) =

1, if (x, y)∈E(G) 0, otherwise.

It is easy to see thatf is anα-admissible mapping and also, f is an α-ψ-Suzuki type rational contraction.

From (i), there exists an x0 ∈ X such that (x0, f x0) ∈ E(G), that is, α(x0, f x0) ≥ 1. Hence, all the conditions of Theorem 2.2 are satisfied and hence,f has a fixed point.

Definition 3.4. Let (X, d) be a b-metric space endowed with a graph G. We say that a self-mapping f :X →X is a G-Suzuki type rational Geraghty contractive mapping if f preserves the edges of G, that is,

(x, y)∈E(G) =⇒(f x, f y)∈E(G), for all x, y∈X and T decreases the weights of the edges of Gin the following way:

1

2sd(x, f x)≤d(x, y) =⇒sd(f x, f y)≤β(M(x, y))M(x, y) (3.2) where,

M(x, y) = max

d(x, y), d(x, f x)d(y, f y)

1 +s[d(x, y) +d(x, f y)], d(x, f y)d(x, y) 1 +s[d(x, f x) +d(y, f y)]

for all (x, y)∈E(G).

Similarly, using Theorem 3.3, we can prove the following theorem.

Theorem 3.5. Let (X, d) be a b-complete b-metric space (with parameter s > 1). Let f be a triangular α-admissible mapping which is a G-Suzuki type rational Geraghty contractive mapping. Also, suppose that the following assertions hold:

(i) there exists x0∈X such that (x0, f x0)∈E(G);

(ii) for any sequence {xn} in X with(xn, xn+1)∈E(G) for alln∈N∪ {0} such that xn→x as n→ ∞, we have (xn, x)∈E(G) for all n∈N∪ {0}.

Then,T has a fixed point.

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4. Contractive mappings on ordered b-metric spaces

Definition 4.1. Let (X, d,) be a partially ordered b-metric space. We say that a self-mappingf :X→X is an orderedψ-Suzuki type rational contractive mapping if

1

2sd(x, f x)≤d(x, y) =⇒d(f x, f y)≤ψ(M(x, y)) (4.1) where,

M(x, y) = max

d(x, y), d(x, f x)d(y, f y)

1 +s[d(x, y) +d(x, f y)], d(x, f y)d(x, y) 1 +s[d(x, f x) +d(y, f y)]

.

for all x, y∈X whitxy.

Theorem 4.2. Let (X, d,)be an ordered b-completeb-metric space. Assume thatf :X→X is an ordered ψ-Suzuki type rational contractive mapping such that the following conditions hold:

(i) there exists an element x0∈X such that x0 f x0; (ii) f is an increasing mapping;

(ii) for any sequence {xn} in X with xnxn+1 for alln∈N∪ {0} such thatxn→x as n→ ∞, we have xnx for alln∈N∪ {0}.

ThenT has a fixed point.

Proof. Define α:X×X→[0,+∞) by

α(x, y) =

1, ifxy 0, otherwise.

Similarly, using Theorem 2.2, we can prove the following theorem.

Theorem 4.3. Let (X, d,) be an ordered b-complete b-metric space (with parameter s > 1). Let T be a non-decreasing ordered Suzuki type rational Geraghty contractive mapping, that is, there exists β ∈ F such that,

1

2sd(x, f x)≤d(x, y) =⇒sd(f x, f y)≤β(M(x, y))M(x, y) (4.2) for all comparable elements x, y∈X, where

M(x, y)

= max

d(x, y),d(x, f x)d(x, f y) +d(y, f y)d(y, f x)

1 +s[d(x, y) +d(f x, f y)] ,d(x, f x)d(x, f y) +d(y, f y)d(y, f x) 1 +d(x, f y) +d(y, f x)

.

Also, suppose that the following assertions hold:

(i) there exists x0∈X such that x0 f x0;

(ii) for any sequence {xn} in X with xnxn+1 for alln∈N∪ {0} such thatxn→x as n→ ∞, we have xnx for alln∈N∪ {0}.

Then,T has a fixed point.

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Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University. The authors, therefore, acknowledge with thanks to DSR for technical and financial support.

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