Fixed Point Theorems for Ciric's and Generalized Contractions in b-Metric Spaces

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ISSN 2291-8639

Volume 7, Number 1 (2015), 70-78 http://www.etamaths.com

FIXED POINT THEOREMS FOR CIRIC’S AND GENERALIZED

CONTRACTIONS IN b-METRIC SPACES

MUHAMMAD SARWAR∗_{AND MUJEEB UR RAHMAN}

Abstract. In this article we obtainedb-metric variant of common fixed point results for Ciric’s and generalized contractions. We have also proved some fixed point results for rational contractive type conditions in the context ofb-metric space. A particular example is also given in the support of our established result regarding Ciric’s type contraction.

1. Introduction

Fixed point theory is one of the most important topic in the development of non linear analysis. In this area, the first important and significant result was proved by Banach in 1922 for contraction mapping in complete metric space. A comprehensive literature and generalization of the Banach’s contraction theorem can be found in [7] and [11]. Some problems particularly the problem of the convergence of measurable functions with respect to measure leads Czerwik (see [5],[6]) to the generalization of metric space and introduce the concept ofb-metric space After Czerwik([5], [6]) many papers have been published containing fixed point results onb-metric spaces for single value and multivalued functions(see[1], [2], [10]).

In this paper we proved b-metric variant of fixed point results for Ciric’s and generalized contraction. We also studied some results involving rational contractive type condition.

Through out the paper_R+ _{will represent the set of non negative real numbers.}

2. Preliminaries

Definition 1. [8]. LetX be a non-empty set and letd:X×X →_R+_{be a function}

satisfying the conditions, d1)d(x, y) = 0 ⇔ x=y;

d2)d(x, y) =d(y, x);

d3)d(x, y)≤d(x, z) +d(z, y) for allx, y, z∈X. Thendit is called metric onX,

and the pair (X, d) is called metric space.

Definition 2. [6]. LetX be a non empty set, letk ≥1 be a real number then a mappingb:X×X→R+is calledb-metric if∀x, y, z∈X, the following conditions are satisfied:

b1)b(x, y) = 0 ⇔ x=y;

2010Mathematics Subject Classification. 47H10, 54H25, 55M20.

Key words and phrases. Completeb-metric space, contraction mapping, self mappings, Cauchy sequence, fixed point.

c

2015 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

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b2)b(x, y) =b(y, x);

b3)b(x, y)≤k[b(x, z) +b(z, y)].And the pair (X, b) is calledb-metric space.

It is clear from the definition ofb-metric that every metric space isb-metric for k= 1, but the converse is not true as clear from the following example.

Example 2.1. [2]. Let X={0,1,2}.Definedb:X×X →R+ as follows

b(0,0) = b(1,1) = b(2,2) = 0, b(1,2) = b(2,1) = b(0,1) = b(1,0) = 1, b(2,0) = b(0,2) =m≥2 fork= m₂ wherem≥2

the function defined above is ab-metric space but not a metric form >2.

For more examples ofb-metric space (see [2], [10]).

In our main work we will use the following definitions which can be found in [2] and [10].

Definition 3. A sequence{xn}in b-metric space (X, b) is called Cauchy sequence if for > 0 there exist a positive integer N such that for m, n ≥ N we have b(xm, xn)< .

Definition 4. A sequence{xn}is called convergent inb-metric space (X, b) if for > 0 and n ≥ N we have b(xn, x) < where x is called the limit point of the sequence{xn}.

Definition 5. A b-metric space (X, b) is said to be complete if every Cauchy sequence inX converge to a point ofX.

Definition 6. [4]. Let (X, d) be a metric space, a self mapping T : X → X is called generalized contraction if and only for all x, y∈ X, there exist c1, c2, c3, c4

such that sup{c1+c2+c3+ 2c4}<1 and

d(T x, T y)≤c1.d(x, y) +c2.d(x, T x) +c3.d(y, T y) +c4.[d(x, T y) +d(y, T x)]

Definition 7. [11]. Let (X, d) be a metric space, a self mapping T : X →X is called Ciric’s type contraction if and only if for allx, y∈X, there existh <1 and,

d(T x, T y)≤h.max

d(x, y), d(x, T x), d(y, T y),d(x, T y) +d(y, T x) 2

Definition 8. [4]. Let (X, d) be a metric space, a self mapping T : X → X is called quasi contraction if and only if for allx, y∈X, there existh <1 and,

d(T x, T y)≤h.max

d(x, y), d(x, T x), d(y, T y), d(x, T y) +d(y, T x)

Remark.

• In [4] the author shown by an example that every quasi contraction need not to be generalized contraction in metric spaces.

Theorem 2.1. [10]. Let (X, b) be a complete b- metric space with k ≥1 and let T :X →X be a contraction with α∈[0,1) andkα <1 thenT has a unique fixed point inX.

Definition 9. [3]. Let (X, d) be a metric space a mapping T :X → X is called weak contraction if there exist constantsα∈(0,1) and someβ≥0 such that

(1) d(T x, T y)≤α.d(x, y) +β.d(y, T x)

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Due to the symmetry of distance the weak contraction clearly include the fol-lowing

(2) d(T x, T y)≤α.d(x, y) +β.d(x, T y)

for allx, y∈X.

Therefore to check the weak contraction of the mapping, we have to check both (1) and (2).

Remarks.

• It is clear that every contraction in metric space is necessarily weak con-traction but the converse is not always true(see[3]).

• Unlike metric space every mapping satisfying the contractive condition in b-metric space need not to be weak contraction necessarily(see[10]).

On the other hand Khan[9] proved the following fixed point result for complete metric spaces.

Theorem 2.2. Let (X, d) be a complete metric space and T be a self mapping on X satisfying the following condition

d(T x, T y)≤µ.d(x, T x).d(x, T y) +d(y, T y).d(y, T x) d(x, T y) +d(y, T x)

∀ x, y∈X andµ∈[0,1), thenT has a unique fixed point.

3. Main Results

Lemma 1. Let(X, b)be ab-metric space and{xn} be a sequence inb-metric space such that

(3) b(xn, xn+1)≤α.b(xn−1, xn)

for n = 1,2,3, ... and 0 ≤ αk < 1, α∈ [0,1), and k is defined in b-metric space then{xn} is a Cauchy sequence in X.

Proof. Letn, m∈N andm > nwe have

b(xn, xm)≤k[b(xn, xn+1) +b(xn+1, xn+2)...]

≤kb(xn, xn+1) +k2[b(xn+1, xn+2) +b(xn+2, xn+3)...]

≤kb(xn, xn+1) +k2b(xn+1, xn+2) +k3b(xn+2, xn+3)...

Now using (3) we have

b(xn, xm)≤kαnb(x0, x1) +k2αn+1b(x0, x1) +k3αn+2b(x0, x1) +...

≤(1 +kα+ (kα)2+...)kαnb(x0, x1)

≤kαn

₁

1−kα

b(x0, x1).

Sincekα <1 therefore taking limitm, n→ ∞we have

lim

m,n→∞b(xn, xm) = 0.

Hence{xn}is a Cauchy sequence inb-metric spaceX.

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Theorem 3.1. Let (X, b) be a completeb-metric space with coefficient k≥1 and T be a self mapping T :X →X satisfying the condition

(4) b(T x, T y)≤α.b(x, y) +β.b(x, T x) +γ.b(y, T y) +µ.[b(x, T y) +b(y, T x)]

∀ x, y∈X, whereα, β, γ, µ≥0, with

(5) kα+kβ+γ+ (k2+k)µ <1

thenT has a unique fixed point.

Proof. Letx0 be arbitrary inX we define a sequence{xn} inX by the rule

x0, x1=T x0, x2=T x1, ..., xn+1=T xn,

consider

b(xn, xn+1) =b(T xn−1, T xn)

by using condition (4) we have

b(xn, xn+1)≤α.b(xn−1, xn)+β.b(xn−1, xn)+γ.b(xn, xn+1)+µ.[b(xn−1, xn+1)+b(xn, xn)]

≤α.b(xn−1, xn) +β.b(xn−1, xn) +γ.b(xn, xn+1) +µ.k[b(xn−1, xn) +b(xn, xn+1]

So

b(xn, xn+1)≤

(α+β+kµ)

(1−(γ+kµ))b(xn−1, xn)

≤λb(xn−1, xn)

where

λ= α+β+kµ 1−(γ+kµ)

From (5) it is clear thatλ <1/k.

Now from Lemma 1 we can say that {xn} is a Cauchy sequence in complete b-metric space, so there exist u ∈X such that limn→∞xn =u. Now we have to

show thatuis the fixed point ofT for this consider,

b(T u, T xn)≤α.b(u, xn) +β.b(u, T u) +γ.b(xn, T xn) +µ.[b(u, T xn) +b(xn, T u)]

≤α.b(u, xn) +β.b(u, T u) +γ.b(xn, xn+1) +µ.[b(u, xn+1) +b(xn, T u)].

Taking limitn→ ∞we have

b(T u, u)≤α.b(u, u) +β.b(u, T u) +γ.b(u, u) +µ.[b(u, u) +b(u, T u)]

≤(β+µ).b(T u, u)

the above inequality is possible only ifb(T u, u) = 0⇒T u=u. Henceuis the fixed point ofT.

Uniqueness. Letu6=v be two fixed points ofT then

b(u, v) =b(T u, T v)≤α.b(u, v) +β.b(u, T u) +γ.b(v, T v) +µ.[b(u, T v) +b(v, T u)]

= (α+ 2µ).b(u, v)

sinceu, v are fixed points of T so finally we get using (5) the above inequality is possible only ifb(u, v) = 0⇒u=v.

Hence fixed point ofT is unique inX.

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Corollary 3.1. Let (X, b) be a completeb-metric space with coefficientk≥1 and T be a self mapping T :X →X satisfying the condition

b(T x, T y)≤α.b(x, y) +β.b(x, T x) +γ.b(y, T y)

∀x, y∈X, whereα, β, γ≥0withkα+kβ+γ <1,thenT has a unique fixed point.

Proof. Puttingµ= 0 in Theorem(3.1) we get the required result easily.

b(T x, T y)≤α.b(x, y) +β.b(x, T x)

∀ x, y∈X, whereα, β≥0 withkα+kβ <1, thenT has a unique fixed point.

Proof. By puttingµ=γ= 0 in Theorem(3.1) we get the required result.

b(T x, T y)≤α.b(x, y)

∀ x, y∈X, whereα≥0 withkα <1 thenT has a unique fixed point.

Proof. By puttingβ =γ=µ= 0 in Theorem(3.1) we get the require result.

Remark.

• Corollary (3.3) is the result of [10].

Now we present the modified form of Khan’s Theorem[9] in the context ofb-metric spaces.

(6) b(T x, T y)≤β.b(x, y) +µ.b(x, T x).b(x, T y) +b(y, T y).b(y, T x) b(x, T y) +b(y, T x)

∀ x, y∈X andβ, µ≥0,b(x, T y) +b(y, T x)6= 0 with k(β+µ)<1, thenT has a unique fixed point.

Proof. Letx0 be arbitrary inX, we define a sequence{xn}by the rule,

x0, x1=T x0, x2=T x1, ..., xn+1=T xn for all n∈N

Now to show that{xn}is a Cauchy sequence in X then consider,

From equation (6) we have

b(xn, xn+1)

≤ β.b(xn−1, xn) +µ.

b(xn−1, T xn−1).b(xn−1, T xn) +b(xn, T xn).b(xn, T xn−1)

b(xn−1, T xn) +b(xn, T xn−1)

≤ β.b(xn−1, xn) +µ.

b(xn−1, xn).b(xn−1, xn+1) +b(xn, xn+1).b(xn, xn) b(xn−1, xn+1) +b(xn, xn)

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Since β+µ < 1_k, therefore by Lemma 1 {xn} is a Cauchy sequence in complete b-metric space X, so there must exist u∈X such that limn→∞xn=u.

Now to show thatuis the fixed point ofT for this consider,

b(T xn, T u) ≤ β.b(xn, u) +µ.

b(xn, T xn).b(xn, T u) +b(u, T u).b(u, T xn) b(xn, T u) +b(u, T xn)

≤ β.b(xn, u) +µ.b(xn, xn+1).b(xn, T u) +b(u, T u).b(u, xn+1) b(xn, T u) +b(u, xn+1)

taking limitn→ ∞we have,

β.b(xn, u) +µ.b(xn, xn+1).b(xn, T u) +b(u, T u).b(u, xn+1) b(xn, T u) +b(u, xn+1)

→0.

Hence

b(u, T u) = 0 ⇒ T u=u

Thusuis the fixed point ofT.

Uniqueness. Letu, v are two distinct fixed points ofT, foru6=v consider,

b(u, v) =b(T u, T v)≤β.b(u, v) +µ.b(u, T u).b(u, T v) +b(v, T v).b(v, T u) b(u, T v) +b(v, T u)

sinceu, v are fixed points ofT we get

b(u, v) =b(T u, T v)≤β.b(u, v) +µ.b(u, u).b(u, v) +b(v, v).b(v, u)

b(u, v) +b(v, u) =β.b(u, v)

using the restriction in the theorem the above inequality is possible only if,

b(u, v) = 0 ⇒ u=v.

Therefore fixed point of T is unique. This complete the proof of the theorem.

Remark.

• In order to hold Khan’s theorem inb-metric space we made no restriction.

(7) b(T x, T y)≤α. b(x, y) +β. b(y, T y)[1 +b(x, T x)] 1 +b(x, y) +γ.

b(y, T y) +b(y, T x) 1 +b(y, T y).b(y, T x)

∀ x, y∈X, whereα, β, γ≥0 and kα+β+γ <1 ThenT has a unique fixed point.

Proof. Letx0 be arbitrary inX, we define a sequence{xn}by the rule,

x0, x1=T x0, x2=T x1, ..., xn+1=T xn for all n∈N

Now to show that{xn}is a Cauchy sequence in X then consider,

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From equation (7) we have

b(xn, xn+1)

≤ α. b(xn−1, xn) +β.

b(xn, xn+1)[1 +b(xn−1, xn)]

1 +b(xn−1, xn)

+γ. b(xn, xn+1) +b(xn, xn) 1 +b(xn, xn+1).b(xn, xn)

≤ α. b(xn−1, xn) +β.

b(xn, xn+1)[1 +b(xn−1, xn)]

1 +b(xn−1, xn)

+γ. b(xn, xn+1)

Therefore

b(xn, xn+1) ≤

α

1−(β+γ). b(xn−1, xn) = h. b(xn−1, xn)

whereh= ₁₋₍α_β₊_γ₎ withh < 1_k, because kα+β+γ <1 similarly we have

b(xn, xn+1)≤h2. b(xn−2, xn−1)

continuing the same process we get

b(xn, xn+1)≤hn. b(x0, x1)

since 0≤h <1 sohn_→_{0 as}_n_{→ ∞}_{, which shows that}_{x

n}is a Cauchy sequence in completeb-metric space so there existz∈X such thatxn →zas n→ ∞ Now to show thatzis the fixed point ofT for this consider,

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b(T xn, T z)≤α. b(xn, z) +β. b(z, T z)[1 +b(xn, T xn)] 1 +b(xn, z)) +γ.

b(z, T z) +b(z, T xn) 1 +b(z, T z).b(z, T xn)

From the construction it is clear that T xn = xn+1 and also {xn} is a Cauchy

sequence converges toz. Therefore taking limitn→ ∞equation (8) become

b(z, T z)≤(β+γ)b(z, T z)

which is possible only ifb(z, T z) = 0,thusT z=z. Hencez is the fixed point ofT.

Uniqueness. Suppose thatT has two fixed pointszandwforz6=wConsider,

b(z, w) =b(T z, T w)≤α. b(z, w)+β. b(w, T w)[1 +b(z, T z)] 1 +b(z, w) +γ.

b(w, T w) +b(w, T z) 1 +b(w.T w).b(w, T z)

So the above inequality become

b(z, w)≤(α+γ). b(z, w)

the above inequality is possible only ifb(z, w) = 0 ⇒ z=w. Hence fixed point

ofT is unique.

Our next theorem is aboutb-metric variant of Ciric’s type contraction.

(9) b(T x, T y)≤h.max

b(x, y), b(x, T x), b(y, T y), 1

2k[b(x, T y) +b(y, T x)]

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Proof. Letx0 be arbitrary inX we define a sequence{xn} inX by the rule

x0, x1=T x0, x2=T x1, ..., xn+1=T xn.

Consider

using (11) we have b(xn, xn+1)

≤h.max

b(xn−1, xn), b(xn−1, T xn−1), b(xn, T xn),

1

2k[b(xn−1, T xn)+b(xn, T xn−1)]

=h.max

b(xn−1, xn), b(xn−1, xn), b(xn, xn+1),

1

2k[b(xn−1, xn+1) +b(xn, xn)]

≤h.max

b(xn−1, xn), b(xn−1, xn), b(xn, xn+1),

1

2[b(xn−1, xn) +b(xn, xn+1)]

(10) =h.max

b(xn−1, xn), b(xn, xn+1),

1

2[b(xn−1, xn) +b(xn, xn+1)]

If

b(xn−1, xn)< b(xn, xn+1).

Then

b(xn−1, xn)<

1

2[b(xn−1, xn) +b(xn, xn+1)]< b(xn, xn+1). So (10) implies that

b(xn, xn+1≤h.b(xn, xn+1)

which is impossible since h < 1 for the same reason we also ignored the term b(xn, xn+1) thus (10) become

b(xn, xn+1)≤h.b(xn−1, xn).

By use of Lemma(1) we get that{xn} is a Cauchy sequence in completeb-metric spaceX so there existu∈X such that limn→∞xn =u, now we have to show that

uis the fixed point ofT for this consider

b(T u, T xn)≤h.max

b(u, xn), b(u, T u), b(xn, T xn), 1

2k[b(u, T xn) +b(xn, T u)]

≤h.max

b(u, xn), b(u, T u), b(xn, xn+1),

1

2k[b(u, xn+1) +b(xn, T u)]

.

Taking limitn→ ∞we have

b(T u, u)≤h.max

b(u, T u), 1

2kb(u, T u)

≤h.b(T u, u)

the above inequality is possible only ifb(T u, u) = 0 ⇒ T u=u. Henceuis the fixed point ofT.

Uniqueness. Letu6=v be two fixed points ofT then

b(u, v) =b(T u, T v)≤h.max

b(u, v), b(u, T u), b(v, T v), 1

2k[b(u, T v) +b(v, T u)]

sinceu, v are the fixed points ofT so finally we have

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the above inequality is possible only ifb(u, v) = 0 ⇒ u=v.

Therefore fixed point ofT inX is unique.

The above theorem produce the following corollary fork= 1.

Corollary 3.4. Let (X, b)be a complete metric space with coefficient and T be a self mappingT :X →X satisfying the condition

(11) b(T x, T y)≤h.max

b(x, y), b(x, T x), b(y, T y),1

2[b(x, T y) +b(y, T x)]

for allx, y∈X, whereh∈[0,1), thenT has a unique fixed point.

Remark. Corollary 3.4 is the result of B. E. Rohades [11].

Example 3.1. LetX = [0,1] andb(x, y) =|x−y|2 _{be a}_b_{-metric with coefficient}

k= 2 ∀x, y∈X, we define the mappingT by

T x= 2

3 if x∈[0,1) and T1 = 0

thenT satisfy all the conditions of the Theorem(3.4) forh∈[4₉,1₂) havingx= 2₃ is its unique fixed point inX.

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Department of Mathematics, University of Malakand, Chakdara Dir(L), Pakistan

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