Some fixed point theorems in complete partial b-metric spaces

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Adv. Fixed Point Theory, 4 (2014), No. 3, 444-461 ISSN: 1927-6303

SOME FIXED POINT THEOREMS IN COMPLETE PARTIAL B-METRIC SPACES

HOSSEIN PIRI∗, HOJJAT AFSHARI

Department of Mathematics, Faculty of Basic Sciences, University of Bonab, Bonab 5551-761167, Iran

Copyright c2014 Piri and Afshari. This is an open access article distributed under the Creative Commons Attribution License, which

permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.In this paper, we prove some fixed point theorems in the framework of complete partialb-metric spaces. The article also includes an example which shows the validity of our results.

Keywords:fixed point, partialb-metric space, contraction. 2010 AMS Subject Classification:47H10, 55M20.

1. Introduction and preliminaries

Fixed point theory is one of the most important tools in development of mathematics because it plays a essential role in applications of many branches of mathematics. For this reason sever-al researchers studied the fixed point of contractive maps (see for example [10] and references therein). In 1992, Polish mathematician Banach [4] proved a very important result regarding a contraction mapping, known as Banach contraction principle. It is one of the fundamental result in fixed point theory. Due to its importance, several authors have obtained many inter-esting extensions and generalizations of metric spaces and Banach contraction principle; see

∗_{Corresponding author}

E-mail address: [email protected] Received May 3, 2014

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[14, 25, 26, 27, 29, 30, 31, 32, 33, 34] and references therein. In this sequel, in 1989, Bakhtin [5] introduced the concept ofb-metric spaces and presented the contraction mapping in b-metric spaces that is generalization of the Banach contraction principle in metric spaces (see also Cz-erwik [9]). After that, fixed point results in b-metric spaces were studied by several researchers; see [3, 6, 20, 19, 13, 21, 11, 7, 22] and references therein.

On the other hand, Matthews [15, 16] introduced the notion of a partial metric space which is a generalization of usual metric space. Also he generalized the Banach contraction principle in the context of complete partial metric spaces. Recently, many researchers have focused on partial metric spaces and obtained many useful fixed point results in these spaces (see [28, 18, 24, 3, 1, 12, 13] and references therein).

Very recently, Shukla [23] introduced partial b-metric spaces as a generalization of both b-metric spaces and partial b-metric spaces. Moreover, he proved Banach contraction principle as well as the Kannan type fixed point theorem in partialb-metric spaces.

In this paper, motivated and inspired by ideas of some recent papers such as Mustafa et al. [17], Suzuki [25], Wong [35] and Shukla [23], we obtain some fixed point theorems in partial b-metric spaces. Our result is the generalization of the result announced by Suzuki [25], Wong [35] and Shukla [23] and some others.

Throughout this paper,_R, _R+ and_N denote the set of real numbers, the set of nonnegative real numbers and the set of positive integers, respectively.

Definition 1.1. [5] Let X be a nonempty set and s≥1 be a given real number. A mapping d: X×X→_R+_{is said to be a b-metric if for all}_x_,_y_,_z_∈_X _{the following conditions are satisfied:}

(bM₁) d(x,y) =0 if and only ifx=y;

(bM₂) d(x,y) =d(y,x);

(bM₃) d(x,z)≤s[d(x,y) +d(y,z)].

In this case, the pair(X,d)is called ab-metric space (with constant s).

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b-metric space instead of a metric space is meaningful since there existb-metric spaces which are not metric spaces.

Example 1.1. LetX = [0,∞) andd: X×X →_R+ defined byd(x,y) =|x−y|p_{, where} _p_{is a} real number such that p>1. Letx,y,z∈X. By takingu=x−zandv=z−y, we have

|x−y|p₌_|_u₊_v_|p_≤₍_|_u_|₊_|_v_|₎p

≤(2 max{|u|,|v|})p ≤2p(|x−z|p+|z−y|p).

Which implies that d(x,y)≤2p[d(x,z) +d(z,y)]. Therefore (X,d) is a b−metric space with coefficient 2p. On the other hand, forx>z>y, we have

|x−y|p=|u+v|p= (u+v)p>up+vp= (x−z)p+ (z−y)p=|x−z|p+|z−y|p,

which implies thatd(x,y)>d(x,z) +d(z,y).Therefore(X,d)is not a metric space.

Definition 1.2.[15] Let X be a nonempty set. A mappingp: X×X→_R+_{is said to be a partial}

metric onX if for allx,y,z∈X the following conditions are satisfied:

(p₁) x=yif and only if p(x,x) =p(x,y) =p(y,y);

(p2) p(x,x)≤ p(x,y);

(p₃) p(x,y) = p(y,x);

(p4) p(x,z)≤ p(x,y) +p(y,z)−p(y,y).

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

The most basic example of a partial metric space is the pair(_R+,p), wherep(x,y) =max{x,y}

for allx,y∈_R+. For more examples of partial metric spaces, we refer to [28, 18, 24, 3, 1, 12, 13].

Definition 1.3. [23] Let X be a nonempty set and s≥1 be a given real number. A mapping p_b: X×X→_R+is said to be a partial b-metric onXif for allx,y,z∈Xthe following conditions are satisfied:

(p_b

1) x=yif and only if pb(x,x) =pb(x,y) =pb(y,y);

(p_b

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(p_b

3) pb(x,y) =pb(y,x);

(p_b

4) pb(x,z)≤s[pb(x,y) +pb(y,z)]−pb(y,y).

In this case, the pair(X,p_b)is called a partial b-metric space (with constant s).

In a partialb-metric space(X,p_b), ifx,y∈X and p_b(x,y) =0, thenx=y, but the converse may not be true. Obviously every partial metric space is a partial b-metric space with the coefficients=1 and everyb-metric space is a partialb-metric space with the same coefficient and zero self-distance. The following example, shows that the converse of these facts need not hold.

Example 1.2. LetX ={0,1,2}and define p_b: X×X →_R+_{as follows:}

p_b(0,0) =0, p_b(1,1) =4, p_b(0,1) =p_b(1,0) =8,

p_b(2,2) =2, p_b(0,2) = p_b(2,0) =5, p_b(1,2) =p_b(2,1) =4.

Because p_b(2,2) =26=0. Thus, condition (bM₁) is not established, therefore (X,p_b) is not b-metric space. Since

8=p_b(o,1)>p_b(o,2) +p_b(2,1)−p_b(2,2) =5+4−2=7,

thus, condition (p4)is not established, therefore, (X,pb) is not partial metric space. But, we

know(X,p_b)is a partialb-metric spaces with constants≥2. For more example, we refer to Shukla [23].

Definition 1.4. [23] Let(X,p_b)be a partial b-metric space with constants, and let{xn}∞_n₌₁be a sequence in X andx∈X. Then:

(i) The sequence{xn}_n∞₌₁is said to converge tox∈X, if limn→∞pb(xn,x) =pb(x,x);

(ii) The sequence{x_n}∞

n=1 is said to be Cauchy in(X,pb), if limn,m→∞pb(xn,xm)exists and

is finite;

(iii) (X,p_b)is said to be complete if for every Cauchy sequence {x_n}∞

n=1 in X there exists x∈X such that

lim n,m→∞

p_b(xn,xm) = lim n→∞

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2. Main results

Theorem 2.1. Let(X,p_b) be a complete partial b-metric space with constant s≥1 and let T be a self mapping on X . Suppose that L∈[0,∞)and there exist functionsαi,i=1,2,3,4,5,6of

(0,∞)into[0,∞)such that

(A) eachαiis upper semicontinuous from the right;

(B) α1(t) +α2(t) +α3(t) +α4(t) +α5(t) +Lα6(t)<_2s12t, for all t>0,

(C) for any distinct x,y∈X ,

1

2spb(x,T x)<pb(x,y)⇒

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p_b(x,y)p_b(T x,Ty)≤α1(pb(x,y))pb(x,y) +α2(pb(x,y))pb(x,Ty)

+α3(pb(x,y))pb(T x,y) +α4(pb(x,y))pb(x,T x)

+α5(pb(x,y))d(y,Ty)

+Lα6(pb(x,y))min{pb(x,y),pb(x,Ty),pb(T x,y)}.

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Then T has a unique fixed point.

Proof. Let us first show that if fixed points ofT exists, then it is unique. Letx,y∈X be two distinct fixed points of T, that is,T x=x6=y=Ty. Therefore p_b(x,y)>0. If p_b(x,x) =0, we have 0= _2s1 p_b(x,x) = _2s1p_b(x,T x)< p_b(x,y).If p_b(x,x)>0. From p_b

2, we get 1

2spb(x,T x) =

1

2spb(x,x)< pb(x,x)≤ pb(x,y).So from pb₂, (B) and (C), we obtain

p_b(x,y)2=p_b(x,y)p_b(T x,Ty)

≤α1(pb(x,y))d(x,y) +α2(pb(x,y))pb(x,Ty)

+α3(pb(x,y))d(T x,y) +α4(pb(x,y))pb(x,T x)

+α5(pb(x,y))pb(y,Ty) +α6(pb(x,y))min{pb(x,y),pb(x,Ty),pb(T x,y)}

≤α1(pb(x,y))pb(x,y) +α2(pb(x,y))pb(x,Ty)

+α3(pb(x,y))pb(T x,y) +α4(pb(x,y))pb(x,T x)

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=α1(pb(x,y))pb(x,y) +α2(pb(x,y))pb(x,y)

+α3(pb(x,y))pb(x,y) +α4(pb(x,y))pb(x,x)

+α5(pb(x,y))pb(y,y) +Lα6(pb(x,y))pb(x,y)

≤α1(pb(x,y))pb(x,y) +α2(pb(x,y))pb(x,y)

+α3(pb(x,y))pb(x,y) +α4(pb(x,y))pb(x,y)

+α5(pb(x,y))pb(x,y) +Lα6(pb(x,y))pb(x,y)

= [α1(pb(x,y)) +α2(pb(x,y)) +α3(pb(x,y))

+α4(pb(x,y)) +α5(pb(x,y)) +Lα6(pb(x,y))]pb(x,y)

< 1

2s2pb(x,y)< pb(x,y).

This is a contradiction. Sox=y. Choosex1∈X. Set

x2=T x1, · · ·,xn+1=T xn=Tn+1x1,· · ·,pn=pb(xn,T xn).

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If there existsn∈_Nsuch that pn=0 the proof is complete. So we assume that for everyn∈N, 0<p_n=p_b(x_n,T x_n) =p_b(x_n,x_n₊₁). Therefore, we have

1

2spb(xn,T xn)<pb(xn,T xn), ∀n∈N.

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From assumption of theorem, we have, p_b(x_n,T x_n)p_b(T x_n,T2x_n)

≤α1(pb(xn,T xn))pb(xn,T xn) +α2(pb(xn,T xn))pb(xn,T

2_x n)

+α3(pb(xn,T xn))pb(T xn,T xn) +α4(pb(xn,T xn))pb(xn,T xn)

+α5(pb(xn,T xn))pb(T xn,T

2_x n)

+Lα6(pb(xn,T xn))min{pb(xn,T xn),pb(xn,T

2_x

n),pb(T xn,T xn)}

≤α1(pb(xn,T xn))pb(xn,T xn) +sα2(pb(xn,T xn))pb(xn,T xn)

+sα2(pb(xn,T xn))pb(T xn,T

2_x

n)−α2(pb(xn,T xn))pb(T xn,T xn)

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−α3(pb(xn,T xn))pb(xn,xn) +α4(pb(xn,T xn))pb(xn,T xn)

2_x

n) +Lα6(pb(xn,T xn))pb(xn,T xn)

≤α1(pb(xn,T xn))pb(xn,T xn) +sα2(pb(xn,T xn))pb(xn,T xn)

+sα2(pb(xn,T xn))pb(T xn,T

2_x

n) +sα3(pb(xn,T xn))pb(T xn,xn)

+sα3(pb(xn,T xn))pb(xn,T xn) +α4(pb(xn,T xn))pb(xn,T xn)

2_x

n) +Lα6(pb(xn,T xn))pb(xn,T xn),

and equivalently,

p_b(x_n,T x_n)p_b(T x_n,T2x_n)≤[α1(pb(xn,T xn)) +sα2(pb(xn,T xn)) +2sα3(pb(xn,T xn))

+α4(pb(xn,T xn)) +Lα6(pb(xn,T xn))]pb(xn,T xn)

+ [sα2(pb(xn,T xn)) +α5(pb(xn,T xn))]pb(T xn,T

2_x n).

So from (3), we get

p_np_n₊₁≤[α1(pn) +sα2(pn) +2sα3(pn) +α4(pn) +Lα6(pn)]pn+ [sα2(pn) +α5(pn)]pn+1

and the equivalently

p_n+1≤

α1(pn) +sα2(pn) +2sα3(pn) +α4(pn) +Lα6(pn) p_n−sα2(pn)−α5(pn)

p_n.

(5)

On the other hand, we get from(B)that

α1(t) +2sα2(t) +2sα3(t) +α4(t) +α5(t) +Lα6(t)≤2s[α1(t) +α2(t) +α3(t)

+α4(t) +α5(t) +Lα6(t)]

<2s 1 2s2t=

1

(8)

Therefore, we have

α1(t) +sα2(t) +2sα3(t) +α4(t) +Lα6(t) t−sα2(t)−α5(t)

<1, ∀t>0.

(6)

It follows from (5) and (6) that

p_b(x_n₊₁,T x_n₊₁) = p_n₊₁<p_n=p_b(x_n,T x_n), ∀n∈_N.

(7)

Therefore,{pn}∞_n₌₁is a decreasing sequence of real numbers which is bounded from below. So

{p_n}∞

n=1converges to some pointp∈[0,∞). If p>0. From (6) and (A), we get p= lim

n→∞

p_n=lim sup n→∞

p_n≤lim sup n→∞

α1(pn) +sα2(pn) +2sα3(pn) +α4(pn) +α6(pn) pn−sα2(pn)−α5(pn)

p_n

=α1(p) +sα2(p) +2sα3(p) +α4(p) +α6(p)

p−sα2(p)−α5(p)

p<p.

This is a contradiction. Hence

lim n→∞

p_n= lim n→∞

p_b(x_n,T x_n) =0.

(8)

Now, we claim that

lim n,m→∞

p_b(x_n,x_m) =0 (9)

Arguing by contradiction, we assume that there exist ε >0 and sequences {p(n)}∞_n₌₁ and {q(n)}∞

n=1of natural numbers such that

p(n)>q(n)>n, p_b(x_p₍_n₎,x_q₍_n₎)≥ε, p_b(x_p₍_n₎₋₁,x_q₍_n₎)<ε, ∀n∈_N.

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Therefore, we have

ε≤p_b(x_p(n),xq(n))≤s[pb(xp(n),xp(n)−1) +pb(xp(n)−1,xq(n))]−pb(xp(n)−1,xp(n)−1)

≤sp_b(x_p₍_n₎,x_p₍_n₎₋₁) +sp_b(x_p₍_n₎₋₁,x_q₍_n₎) ≤sp_b(x_p₍_n₎,x_p₍_n₎₋₁) +sε

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Using (8), we get

ε≤lim inf

n→∞

p_b(x_p₍_n₎,x_q₍_n₎)≤lim sup n→∞

p_b(x_p₍_n₎,x_q₍_n₎)≤sε.

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On the other hand, we find from (8) that limn→∞pb(xp(n),T xp(n)) =0.So there existsN∈N

such that _2s1 p_b(x_p₍_n₎,T x_p₍_n₎)<ε,∀n>N,and from (10), we have

1

2spb(xp(n),T xp(n))< pb(xp(n)+1,xq(n)+1) =pb(T xp(n),T xq(n)), ∀n>N.

Therefore, from (C) for everyn>N, we obtain p_b(x_p₍_n₎,x_q₍_n₎)p_b(T x_p₍_n₎,T x_q₍_n₎)

≤α1(pb(xp(n),xq(n)))pb(xp(n),xq(n)) +α2(pb(xp(n),xq(n)))pb(xp(n),T xp(n))

+α3(pb(xp(n),xq(n)))pb(xq(n),T xq(n)) +α4(pb(xp(n),xq(n)))pb(xp(n),T xq(n))

+α5(pb(xp(n),xq(n)))pb(T xp(n),xq(n))

+Lα6(pb(xp(n),xq(n)))min{pb(xp(n),xq(n)),pb(xp(n),T xq(n)),pb(T xp(n),xq(n))}

+α3(pb(xp(n),xq(n)))pb(xq(n),T xq(n)) +α4(pb(xp(n),xq(n)))pb(xp(n),T xq(n))

+α5(pb(xp(n),xq(n)))pb(T xp(n),xq(n)) +Lα6(pb(xp(n),xq(n)))pb(xp(n),xq(n)),

and this is equivalent to

p_b(x_p₍_n₎,x_q₍_n₎)p_b(x_p₍_n₎₊₁,x_q₍_n₎₊₁)

+α3(pb(xp(n),xq(n)))pb(xq(n),T xq(n)) +sα4(pb(xp(n),xq(n)))pb(xp(n),xq(n))

+sα4(pb(xp(n),xq(n)))pb(xq(n),T xq(n))−α4(pb(xp(n),xq(n)))pb(xq(n),xq(n))

+sα5(pb(xp(n),xq(n)))pb(T xp(n),xp(n)) +sα5(pb(xp(n),xq(n)))pb(xp(n),xq(n))

−α5(pb(xp(n),xq(n)))pb(xp(n),xp(n)) +Lα6(pb(xp(n),xq(n)))pb(xp(n),xq(n))

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+α4(pb(xp(n),xq(n)))pb(xq(n),T xq(n)) +α5(pb(xp(n),xq(n)))pb(T xp(n),xp(n))

+α5(pb(xp(n),xq(n)))pb(xp(n),xq(n)) +Lα6(pb(xp(n),xq(n)))pb(xp(n),xq(n)).

It follows that

p_b(x_p₍_n₎,x_q₍_n₎)p_b(x_p₍_n₎₊₁,x_q₍_n₎₊₁)

≤[α1(pb(xp(n),xq(n))) +sα4(pb(xp(n),xq(n)))

+sα5(pb(xp(n),xq(n))) +Lα6(pb(xp(n),xq(n)))]pb(xp(n),xq(n))

+ [α2(pb(xp(n),xq(n))) +sα5(pb(xp(n),xq(n)))]pb(T xp(n),xp(n))

+ [α3(pb(xp(n),xq(n))) +sα4(pb(xp(n),xq(n)))]pb(xq(n),T xq(n)).

Hence from (B), we obtain

p_b(x_p₍_n₎,x_q₍_n₎)p_b(x_p₍_n₎₊₁,x_q₍_n₎₊₁)≤ 1

2s3[pb(xp(n),xq(n)]

2

+ 1

2s2pb(xp(n),xq(n))pb(T xp(n),xp(n))

+ 1

2s2pb(xp(n),xq(n))pb(xq(n),T xq(n)).

(12)

From (8), (10), (11) and (12), we conclude that

ε2≤lim sup

n→∞

[p_b(x_p₍_n₎,x_q₍_n₎)p_b(x_p₍_n₎₊₁,x_q₍_n₎₊₁)]

≤lim sup n→∞

[ 1

2s3(pb(xp(n),xq(n))

2

+ 1

2s3pb(xp(n),xq(n))pb(T xp(n),xp(n))

+ 1

2s3pb(xp(n),xq(n))pb(xq(n),T xq(n))]

≤ 1

2s2(sε) 2₌1

2ε 2_<

ε2.

This is a contradiction. Hence limm,n→∞d(xn,xm) =0. By completeness of(X,pb)there exists

x∈X such that

p_b(x,x) = lim n→∞

p_b(xn,x) = lim n,m→∞

(11)

We shall prove that for everyn∈_N,

(I) 1

2spb(xn,T xn)<pb(xn,x),

or (14)

(II) 1

2spb(T xn,T

2_x

n)<pb(T xn,x).

Arguing by contradiction, we assume that there existsm∈_Nsuch that

1

2spb(xm,T xm)≥pb(xm,x)and

1

2spb(T xm,T

2_x

m)≥ pb(T xm,x).

(15)

Therefore, from (5), (7) and (15), we obtain

p_b(x_m,T x_m)≤s[p_b(x_m,x) +p_b(x,T x_m)]−p_b(x,x) ≤sp_b(xm,x) +spb(x,T xm)

≤s1

2spb(xm,T xm) +s

1

2spb(T xm,T

2_x m)

<1

2pb(xm,T xm) +

1

2pb(xm,T xm) =pb(xm,T xm).

This is a contradiction. Hence (14) holds. From part(I)of (14), we get

p_b(x_n,x)p_b(T x_n,T x)≤α1(pb(xn,x))pb(xn,x) +α2(d(xn,x))pb(xn,T x)

+α3(pb(xn,x))pb(T xn,x) +α4(pb(xn,x))pb(xn,T xn)

+α5(pb(xn,x))d(x,T x)

+Lα6(pb(xn,x))min{pb(xn,x),pb(xn,T x),pb(T xn,x)}

≤α1(pb(xn,x))pb(xn,x) +sα2(pb(xn,x))pb(xn,T xn)

+sα2(pb(xn,x))pb(T xn,T x) +α3(pb(xn,x))pb(T xn,x)

+α4(pb(xn,x))pb(xn,T xn) +sα5(pb(xn,x))pb(x,T xn)

(12)

which is equivalent to

p_b(x_n,x)p_b(T x_n,T x)≤[α1(pb(xn,x)) +Lα6(pb(xn,x))]pb(xn,x)

+ [sα2(pb(xn,x)) +α4(pb(xn,x))]pb(xn,T xn)

+ [sα2(pb(xn,x)) +sα5(pb(xn,x))]pb(T xn,T x)

+ [α3(pb(xn,x))pb(T xn,x) +sα5(pb(xn,x))]pb(x,T xn).

This implies that

p_b(T x_n,T x)≤ α1(pb(xn,x)) +Lα6(pb(xn,x))

p_b(xn,x)−sα2(pb(xn,x))−sα5(pb(xn,x)

p_b(x_n,x)

+ sα2(pb(xn,x)) +α4(pb(xn,x))

p_b(x_n,x)−sα2(pb(xn,x))−sα5(pb(xn,x))

p_b(x_n,T x_n)

+ α3(pb(xn,x)) +α5(pb(xn,x))

p_b(xn,x)−sα2(pb(xn,x))−sα5(pb(xn,x))

p_b(x,T x_n).

(16)

From (6) and (16), we have

p_b(T x_n,T x)≤p_b(x_n,x) +p_b(x_n,T x_n) +p_b(x,T x_n) =p_b(x_n,x) +p_n+p_b(x,x_n₊₁).

Using (8) and (13), we obtain

lim n→∞

p_b(T x_n,T x) =0.

(17)

Since

p_b(x,T x)≤s[p_b(x,T x_n) +p_b(T x_n,T x)]−p_b(T x_n,T x_n) ≤s[p_b(x,T x_n) +p_b(T x_n,T x)]

(13)

we find from (13) and (17) thatx=T x. From part(II)of (14), we get

p_b(x_n,x)p_b(T2x_n,T x)

≤α1(pb(T xn,x))pb(T xn,x) +α2(pb(T xn,x))pb(T xn,T x)

+α3(pb(T xn,x))pb(T

2_x

n,x) +α4(pb(T xn,x))pb(T xn,T

2_x n)

+α5(pb(T xn,x))pb(x,T x)

+Lα6(pb(T xn,x))min{pb(T xn,x),pb(T xn,T x),pb(T

2_x n,x)}

≤α1(pb(T xn,x))pb(T xn,x) +sα2(pb(T xn,x))pb(T xn,T

2_x n)

+sα2(pb(T xn,x))pb(T

2_x

n,T x)−α2(pb(T xn,x))pb(T

2_x

n,T2xn)

+α3(pb(T xn,x))pb(T

2_x

n,x) +α4(pb(T xn,x))pb(T xn,T

2_x n)

+sα5(pb(T xn,x))pb(x,T

2_x

n) +sα5(pb(T xn,x))pb(T

2_x n,T x)

−α5(pb(T xn,x))pb(T

2_x

n,T2xn) +Lα6(pb(T xn,x))pb(T xn,x)

≤α1(pb(T xn,x))pb(T xn,x) +sα2(pb(T xn,x))pb(T xn,T

2_x n)

2_x

n,T x) +α3(pb(T xn,x))pb(T

2_x n,x)

+α4(pb(T xn,x))pb(T xn,T

2_x

n) +sα5(pb(T xn,x))pb(x,T

2_x n)

2_x

n,T x) +Lα6(pb(T xn,x))d(T xn,x),

which is equivalent to

p_b(T xn,x)pb(T

2_x

n,T x)≤[α1(pb(T xn,x)) +Lα6(pb(T xn,x))]pb(T xn,x)

+ [sα2(pb(T xn,x)) +α4(pb(T xn,x))]pb(T xn,T

2_x n)

+ [sα2(pb(T xn,x)) +sα5(pb(T xn,x))]pb(T

2_x n,T x)

+ [α3(pb(T xn,x)) +sα5(pb(T xn,x))]pb(T

(14)

This implies that p_b(T2xn,T x)<

α1(pb(T xn,x)) +Lα6(pb(T xn,x))

p_b(T x_n,x)−sα2(pb(T xn,x))−sα5(pb(T xn,x)

p_b(T xn,x)

+ sα2(pb(T xn,x)) +α4(pb(T xn,x))

p_b(T x_n,x)−sα2(pb(T xn,x))−sα5(pb(T xn,x))

p_b(T x_n,T2x_n)

+ α3(pb(T xn,x)) +sα5(pb(T xn,x))

p_b(T xn,x)−sα2(pb(T xn,x))−sα5(pb(T xn,x))

p_b(x,T2x_n).

(18)

From (6) and (18), we conclude that

0≤ p_b(T2x_n,T x)≤p_b(T x_n,x) +p_b(T x_n,T2x_n) +p_b(x,T2x_n) =p_b(x_n+1,x) +pn+1+pb(x,xn+2).

Hence from (8) and (13), we obtain lim n→∞

p_b(T2xn,T x) =0. (19)

Since

p_b(x,T x)≤s[p_b(x,T2x_n) +p_b(T2x_n,T x)]−p_b(T2x_n,T2x_n) ≤s[p_b(x,T2x_n) +p_b(T2x_n,T x)]

=sp_b(x,x_n₊₂) +p_b(T2x_n,T x),

we obtain from (13) and (19) thatx=T x.

Theorem 2.2.Let (X, b) be a complete partial b-metric space with coefficient s>1and T:X→

X be a mapping satisfying the following condition:

b(T x,Ty)≤λb(x,y),∀x,y∈X,

whereλ ∈[0,1). Then T has a unique fixed point u∈X and b(u,u) =0. Proof.It is sufficient to takeα1(t) =_2s13t,λ =

1

2s3 andα2(t) =α3(t) =α4(t) =α5(t) =α6(t) = 0 in Theorem .

Theorem 2.3.Let (X, b) be a complete partial b-metric space with coefficient s>1and T:X→

X be a mapping satisfying the following condition:

(15)

whereλ ∈[0,1₂)andλ 6=1_s.

Proof.It is sufficient to takeα4(t) =α5(t) = _2s13t,λ= 1

2s3 andα1(t) =α2(t) =α3(t) =α6(t) = 0 in Theorem .

Example 2.1. Put X =_N∪ {m+_n₊1₂ :m,n∈_N}. Let P be a real number such that p>1and d: X×X →_R+ be defined by

p_b(x,y) =max{x,y}p+|x−y|p.

Define a mapping T on X by

T(x) = 





1 x∈_N,

7m+_n₊1₂ x=m+1_n.

Then T satisfies in the assumption of Theorem 2.1.

Proof. It is obviously that(X,p_b)is a complete partial b-metric space with coefficient s=2p, but it is neither a b-metric nor a partial metric space [23] and 1 is a unique fixed point ofT. if n<m, we have

1 2d(m+

1

n+2,T(m+ 1 n+2))

= 1

2d(m+ 1

n+2,7m+ 1 n+2)

= 1

2

max{m+ 1

n+2,7m+ 1

n+2}+|m+ 1

n+2−7m− 1 n+2 |

= 1

2(7m+ 1

n+2+6m)

= 1

2(13m+ 1 n+2) and

d(m+ 1

n+2,n+ 1

n+2) =max{m+ 1 n+2,n+

1

n+2}+|m+ 1

n+2−n− 1 n+2 |

=m+ 1

n+2+m−n

<2m+ 1

n+2.

(16)

Conflict of Interests

The authors declare that there is no conflict of interests.

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