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J. Semigroup Theory Appl. 2019, 2019:6 https://doi.org/10.28919/jsta/3743 ISSN: 2051-2937

COMMON FIXED POINT IN D∗ METRIC SPACES

NEVEEN ALZUDE, ABDALLA TALLAFHA∗

Department of mathematics, The University of Jordan, Amman-Jordan

Copyright c2018 Alzude and Tallafha. 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 shall obtaine a new results conserning fixed point inD∗ Metric Spaces, besides we correct the proves of some results obtaned by, T. Veerapandi and AJI. M Pillai in[35].

Keywords:metric spaces;D∗metric Spaces; semi-linear uniform spaces; contractions; fixed point; common fixed point.

2010 AMS Subject Classification:Primary 54E35, Secondary 41A65.

1.

INTRODUCTION

The most important principle as a source of existence and uniqueness theorem in different

branches of Sciences is Banach contraction principle, since it gives the existence, uniqueness

and the sequence of the successive approximation converges to solution.

The study of metric fixed point theory plays an important role in many important areas

as differential equation, operation research, mathematical economies and other branches, see

[1],[4],[5],[7],[9]−[11],[13]−[17],[19]−[22],[25]−[28],[36]and[37].

Corresponding author

E-mail address: a.tallafha@ju.edu.jo

Received May 9, 2018

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Several generalization of metric spaces were proposed by several mathematicians such as

2-metric spaces, Gahler[8].D-metric spaces, B. C. Dhage[6], G- metric space[18], L. -G. Huaug

at X. Zhang, b- metric spaces,[12].

Recently A.Tallafha and R. Khalil[29],defined a space which is a mixture of analysis and

topology, namely semi-linear uniform space. Semi-linear uniform space is weaker than metric

space and stronger than topological space since. Several authors studied the properties of

semi-linear uniform spaces and fixed point in such spaces, see[2],[3],[23],[24],and[30]−[34].

In 2011, T. Veerapandi and AJI. M Pillai in[35],established some common fixed point

the-orems for contraction and generalized contraction mapping inD∗-metric spaces. In the proves of the main results, T. Veerapandi and AJI. M Pillai found a sequence nameddn∗ that satisfies

dn+1−dn∗≤αnd0, since α ∈[0,1), then αnd0→0. From this they conclude thatdn+1≤dn∗.

This conclusion not correct by the following example.

Example 1. Let xn=1−21n,then xn+1=1−2n1+1, therefor xn+1−xn= 1 2n+1−

1

2n =2n1+1 = 1 2

n

1 2

,soα=d0=21,clearly xn+1≥xn.

In this article, we define an orbit of two and three mapping on aD∗-metric spaceX,also we correct the proves given in[35],under additional condition. Finally we obtained anew results of

common fixed point for contraction mapping inD∗-metric spaces.

2.

D∗METRICSPACES

Definition 1. Let X be a non-empty set. A generalized metric space or (D∗-metric space) on X is a function D∗:X3→[0,∞)that satisfies the following conditions, for each x,y,z,a∈X.

i) D∗(x,y,z)≥0.

ii) D∗(x,y,z) =0if and only if x=y=z.

iii) D∗(x,y,z) =D∗(P{x,y,z}).where P is permutation. iv) D∗(x,y,z)≤D∗(x,y,a) +D∗(a,z,z).

The pair(X,D∗)is called generalized metric space or( D∗-metric space).

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Definition 3. Let(X,D∗)be a D∗-metric space and A⊂X,

i) A subset A of X is said to be D∗−bounded if there exists r>0,such that D∗(x,y,y)<r,

for all x,y∈A.

ii) A sequence{xn}in X converges to x if and only if

D∗(xn,xn,x) =D∗(x,x,xn)→0as n→∞.That is, for eachε>0there exists n0∈N such that D∗(x,x,xn)<ε for all n≥n0.

iii) A sequence{xn}in X is called a Cauchy sequence if for eachε>0there exists n0 ∈N such that D∗(xn,xn,xm)<ε for all n,m≥n0.

iv) the D∗-metric space(X,D∗)is said to be complete if every Cauchy sequence is convergent.

Definition 4. (i)Let x∈X and f is a self mapping then the set O(f,x) ={fn(x):n=0,1,2,3, ..}is called the orbit of x and f.

(ii) Let x∈X and f1,f2 are self mapping then the set O(f1,f2,x) ={x,f1i

f2j(x): j,i= 0,1,2,3, ..}is called the orbit of x,f1and f2

(iii)Let x∈X and f1,f2,f3are self mapping then the set O(f1,f2,f3,x) ={x,f1j f2i f3k(x):

j,i,k=0,1,2,3, ..}is called the orbit of x,f1, f2and f3

Now we want to present a new definition.

Definition 5. A three self mapping f1,f2 and f3 on a D∗-metric space(X,D∗),are said to be

orbitally bounded if there exists x0 ∈X,and a three positive real numbers λ1,λ2andλ3such

that D∗(a,b,fi(x))≤λiD∗(a,b,x),for all x∈O(f1,f2,f3,x0)and{a,b} 6={x0}.

Ifmax{λ1,λ2,λ3}<1,then f1,f2and f3are called orbitally contractive.

In the above definition we assume a,b not booth equal x0, since if a=b=x0, the above condition implies thatx0 is the common fixed point.

One of the interesting properties ofD∗-metric space, are the following lemmas.

Lemma 1. In D∗-metric space D∗(x,y,y) =D∗(x,x,y).

Lemma 2. Let (X,D∗)be a D∗−metric space then limn→∞D∗(xn,yn,zn) =D∗(x,y,z)

when-ever a sequence {(xn,yn,zn)} in X3 converges to a point (x,y,z) in X3.That islimn→∞xn =

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Proof. Let{(xn,yn,zn)} ∈X3,be such that limn→∞xn =x,limn→∞yn =y,

limn→∞zn =z.Then for eachε>0,there existsn1,n2andn3∈N,such that D∗(x,x,xn)< ε

2, for alln≥n1,D∗(y,y,yn)<

ε

2,for all n≥n2 and D∗(z,z,zn)<

ε

2, for all

n≥n3.Letn0=max{n1,n2,n3},then for everyn≥n0by triangle inequality we obtain,

D∗(xn,yn,zn)≤D∗(xn,yn,z) +D∗(z,zn,zn)

≤D∗(x,y,z) +D∗(x,xn,xn) +D∗(y,yn,yn) +D∗(z,zn,zn)

<D∗(x,y,z) +ε

3+

ε

3+

ε

3 =D∗(x,y,z) +ε. ThenD∗(xn,yn,zn)−D∗(x,y,z)<ε.Moreover, D∗(x,y,z)≤D∗(x,y,zn) +D∗(zn,z,z)

≤D∗(zn,yn,zn) +D∗(xn,x,x) +D∗(yn,y,y) +D∗(zn,z,z).

<D∗(xn,yn,zn) +ε

3+

ε

3+

ε

3=D∗(xn,yn,zn) +ε. therefore,D∗(x,y,z)−D∗(xn,yn,zn)<ε,which implies, |D∗(xn,yn,zn)−D∗(x,y,z)|<ε,therefore

limn→∞D∗(xn,yn,zn) =D∗(x,y,z)

One can easly obtained the following Lemmas.

Lemma 3. Let(X,D∗)be a D∗-metric space, then a convergent sequence has a unique limit.

Lemma 4. Let (X,D∗) be a D∗-metric space. Then any convergent sequence in (X,D∗) is a Cauchy sequence .

3.

MAIN RESULTS

In this section, we will present several fixed point results on a completeD∗−metric space.

Theorem 1. Let X be a complete D∗-metric space and f1,f2:X →X be any two maps such that,

D∗(f1(x),f2(y),z)≤αD∗(x,y,z)for all x,y∈O(f1,f2,x0),z∈X and some x0∈X,and0≤

α < 12,then f1,f2have a unique common fixed point.

Proof. Letx0∈X be any aribitrary element defined a sequence{xn}inX by,

xn=

f

1(xn−1), ifnis odd

f2(xn−1), ifnis even

.Now we want to prove that{xn} Cauchy sequence, then we

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Case 1.Ifnis odd, setdn=D∗(xn,xn+1,xn+1),so

dn=D∗(xn,xn+1,xn+1) =D∗(f1(xn−1),f2(xn),xn+1)

≤αD∗(xn−1,xn,xn+1) ≤αD∗(xn−1,xn,xn) +αD∗(xn+1,xn+1,xn).

Hence,dn≤α dn−1+α dn,which impliesdn≤ 1−ααdn−1. letR= α

1−α soR<1 sinceα<

1

2,anddn≤R dn−1.Repeating this process, we obtained,

dn≤Rnd0,i.e dn→0 asn→∞.

To prove that{xn}Cauchy sequence, letn,m∈Nbe such thatn≤m,then

D∗(xn,xn,xm)≤D∗(xn,xn,xn+1) +D∗(xm,xm,xn+1) ≤∑mk=−1n D∗(xk,xk+1,xk+1)≤∑mk=−1n dk

≤∑mk=−1n Rkd0 ≤Rnd0 1−1k

→0 asn→∞.

Case 2: Ifnis even, setdn=D∗(xn,xn+1,xn+1),so

dn=D∗(xn+1,xn,xn+1) =D∗(f1(xn),f2(xn−1),xn+1)

≤α D∗(xn,xn−1,xn+1) ≤αD∗(xn−1,xn,xn) +αD∗(xn+1,xn+1,xn),

which impliesdn≤α dn−1+α dn.As in case 1, we obtain,

D∗(xn,xn,xm)→0 asn→∞.So{xn}is Cauchy sequence but sinceX is completeD∗-metric

space thenxnis converges to some element sayx.

Now we want to prove thatxis fixed point of f1,suppose thatx6= f1(x),thenD∗(f1(x),x,x) = limn→∞D∗(f1(x),f1(x2n−1),x)≤limn→∞αD∗(x,x2n−1,x) =0,

thereforeD∗(f1(x),x,x) =0, soxis fixed point for f1.Similarlyxis fixed point for f2. For uniqueness, suppose there existsy∈X such that, f1(y) = f2(y) =y.

This impliesD∗(x,y,y) =D∗(f1(x),f2(y),y)≤αD∗(x,y,y)< 12D∗(x,y,y),which impliesx=

y.

The next theorems give the same result for orbitally contractive maps under some certain

conditions.

Theorem 2. Let X be a complete D∗-metric space and f1,f2,f3:X →X be any three orbitally

contractive maps that satisfies

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Proof. LetX be a completeD∗-metric space and f1,f2,f3:X →X be any three orbitally con-tractive maps that satisfies,D∗(f1(x),f1(x),f2(y))≤D∗(f1(x),f2(y),f3(z)).Since f1,f2,f3are orbitally contractive maps, then there exists x0 ∈X, and a three positive numbers λ1,λ2and

λ3such thatD∗(a,b,fi(x))≤λiD∗(a,b,x),for allx∈O(f1,f2,f3,x0)and{a,b} 6={x0},and max{λ1,λ2,λ3}=α <1.

Now we want to show

D∗(f1(x),f2(y),f3(z))≤λ1λ2λ3D∗(x,y,z),for allx,y,z∈O(f1,f2,f3,x0).

Letx,y,z∈O(f1,f2,f3,x0),ifx=y=z=x0,thenx0 = f1(x0) = f2(x0) = f3(x0), then the required inequality holds.

Defined a sequence{xn}as follows,

xn=     

   

f1(xn−1),n=1,4,7, ....3k+1

f2(xn−1),n=2,5,8, ....3k+2

f1(xn−1),n=3,6,9, ....3k+3

    

    .

Now we want to prove thatxn is Cauchy sequence, that is D∗(xn,xn,xm)→0 as n,m→∞, with no loss of generality assumem≥n.

sinceD∗(xn,xn,xm) =D∗(xm,xm,xn),then we have the following cases. Case1:xn= f1(xn−1),xm= f2(xm−1),

D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1)

≤D∗(f1(xn−1),f2(xm−1),f3(xn−1))≤αnD∗(x0,x0,f2(xm−n))

≤αnλ2D∗(x0,x0,xm−n)→0 asm,n→∞.

Case 2: xn= f1(xn−1),xm= f3(xm−1),

D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1)

≤D∗(f1(xn−1),f2(xm−1),f3(xn−1)) ≤αnD∗(x0,x0,f2(xm−n))

≤αnλ2D∗(x0,x0,xm−n)→0 asm,n→∞.

Case 3: xn= f2(xn−1),xm= f3(xm−1),

D∗(xn,xn,xm) =D∗(f2(xn−1),f2(xn−1),f3(xm−1)

≤D∗(f1(xn−1),f2(xn−1),f3(xm−1))≤αnD∗(x0,x0,f3(xm−n))

≤αnλ3D∗(x0,x0,xm−n)→0 asm,n→∞.

So{xn}is Cauchy sequence in the completeD∗-metric spaceX, then there existx∈X, such

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Now we want to prove thatxis fixed point of f1and by a similar idea we can show thatxis fixed point of f2and f3.

xis a fixed point for f1.

D∗(f1(x),x,x) =limn→∞D∗(f1(x),x3n+2,x3n+3)

=limn→∞D∗(f1(x),f2(x3n+1),f3(x3n+2)) ≤αlimn→∞D∗(x,x3n+1,x3n+2)

=αD∗(x,x,x).so f1(x) =x.

Now we want to prove uniqueness of the fixed pointx.Assume there exist another common

fixed point y, that is f1(y) = f2(y) = f3(y) =y. Now D∗(x,y,y) =D∗(f1(x),f2(y),f3(y))≤

αD∗(x,y,y)

Sinceα ∈(0,1),thenD∗(x,y,y) =0,hencey=x.

Theorem 3. Let X be a complete D∗-metric space and g,f :X →X be any two maps such that D∗(g f(x),f(x),y)≤αD∗(f(x),x,y), for all x,y∈O(f,g,x0),x0∈X and 0≤α < 12,then

g,f have a unique common fixed point.

Proof. Defined a sequence{xn}inX as follow,

xn=

f

(xn−1), ifnis dd

g(xn−1), ifnis even

,

Setdn=D∗(xn,xn+1,xn+1)for alln=0,1,2, ..., nowd1=D∗(x1,x2,x2) =D∗(f(x0),g f(x0),x2)

≤αD∗(x0,f(x0),x2) =αD∗(x0,x1,x2)

≤αD∗(x0,x1,x1) +αD∗(x1,x2,x2) =αd0+αd1,therefor

d1≤βd0,whereβ =1−αα <1.Also,

d2=D∗(x2,x3,x3) =D∗(g f(x0),f(x2),x3) ≤αD∗(f(x0),x2,x3) =αD∗(x1,x2,x3) ≤αD∗(x1,x2,x2) +αD∗(x2,x3,x3)

=αd1+αd2,thereford2≤βd1.By a similar arguments we have,

dn≤βdn−1≤βnd0→0 asn→∞.

Now we shall prove that{xn}is Cauchy sequence inX.

Letn,m>n0,for somen0N,then

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=∑mk=−1n βn1−1

βd0→0 asn,m→∞,therefore{xn}is aD

Cauchy sequence

in a completeD∗−metric spaceX,hence converges to somex∈X.Since,

D∗(f(x),x,x) =limn→∞D∗(f(x),x2n,x) =limn→∞D∗(f(x),g(x2n−1),x) =limn→∞D∗(f(x),g f(x2n−2),x)≤αlimn→∞D∗(f(x),x2n−2,x) =αlimn→∞D∗(f(x),x2n−2,x) =αlimn→∞D∗(f(x),x,x)

thereforeD∗(g f(x),f(x),x) =D∗(f(x),x,x) =0,which imliesg(x) = f(x) =x.

To prove uniqueness, if possible suppose ther exist x6=y, such that g(x) = f(x) = x and

g(y) = f(y) =y,

thenD∗(x,x,y) =D∗(g f(x),f(x),y)≤αD∗(f(x),x,y) =αD∗(x,x,y).

This impliesα≥1.x=y.

Theorem 4. Let X be a complete D∗-metric space and f,g,h:X→X be any three maps satisfy the following conditions

1) There exists, x0 ∈X,and0<λ<1,such that D∗(a,b,f(x))≤λD∗(a,b,x),for all a,b,x∈ O(f1,f2,f3,x0)and{a,b,x} 6={x0}

2) D∗(hg f(x),g f(x),f(x))≤αD∗(g f(x),f(x),x)for all x∈X and0<α <1,λ <1. 3) For all x,y,z∈O(f,g,h,x0),we have

D∗(f(x),f(x),g(y))≤D∗(f(x),g(y),h(z)),

D∗(f(x),f(x),h(z))≤D∗(f(x),g(y),h(z))

and D∗(g(y),g(y),h(z))≤D∗(f(x),g(y),h(z)).

Then f,g,h have a unique common fixed point.

Proof. Let X be a complete D∗-metric space and f,g,h :X → X be three maps satisfy the above conditions. By (2), there exists, x0 ∈X, and a positive real number λ, such that D∗(a,b,f1(x))≤λD∗(a,b,x),for allx∈O(f,g,h,x0)and{a,b,x} 6={x0}. Define a sequence {xn}inX as follow

xn=

    

   

f(xn−1), n=1,4,7, .. . . .3k+1

g(xn−1), n=2,5,8, , . . .3k+2

h(xn−1), n=3,6,9...3k+3

    

   

To prove{xn}is Cauchy sequence, with no loss of generality assumem≥n

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D∗(xn,xn,xm) =D∗((f(xn−1),f(xn−1),g(xm−1) ≤D∗(h(xn−1),g(xm−1),f(xn−1))

=D∗(hg f(xn−3),g f(xm−2),f(xn−1)) ≤αD∗(g f(xn−3),f(xm−2),(xn−1) ) =αD∗(xn−1,xm−1,xn−1)≤...

≤αnD∗(x0,x0,f(xm−n−1))

≤αn(λ)m−nD∗(x0,x0,f(x0))→0 asn,m→∞

Case 2: xn=g(xn−1),xm=h(xm−1),

D∗(xn,xn,xm) =D∗(g(xn−1),g(xn−1),h(xm−1) ≤D∗(h(xm−1),g(xn−1),f(xn−1))

=D∗(hg f(xm−3),g f(xn−2),f(xn−1)) ≤αD∗(g f(xm−3),f(xn−2),(xn−1) ) =αD∗(xn−1,xm−1,xn−1)≤...

≤αnD∗(x0,f(xm−n−1),x0)

≤αn(λ)m−nD∗(x0,x0,f(x0))→0 asn,m→∞

Cas 3: xn= f(xn−1),xm=h(xm−1),

D∗(xn,xn,xm) =D∗((f(xn−1),f(xn−1),h(xm−1) ≤D∗(h(xm−1),g(xn−1),f(xn−1))

=D∗(hg f(xm−3),g f(xn−2),f(xn−1)) ≤αD∗(g f(xm−3),f(xn−2),(xn−1) )

=αD∗(xn−1,xm−1,xn−1)≤... ≤αnD∗(x0,f(xm−n−1),x0)

≤αn(λ)m−nD∗(x0,x0,f(x0))→0 asn,m→∞.

So{xn}converges to some element sayx∈X.

Now we want to prove thatxis fixed point of f,g,h.

(i)D∗(f(x),x,x) =limn→∞D∗(f(x),x3n−3,x3n−1)

=limn→∞D∗(hg f(x3n−5),g f(x3n−2),f(x)) ≤α limn→∞D∗(x3n−3,x3n−1,x) =αD∗(x,x,x)

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=limn→∞D∗(hg f(x3n−5),g f(x),f(x)) ≤αlimn→∞D∗(g f(x3n−5),f(x),x) =αlimn→∞D∗(x3n−5,x,x))D∗(x,x,x)) (iii)D∗(x,x,h(x)) =D∗(f(x),g f(x),hg f(x))

≤αD∗(x,f(x),f g(x))

=αD∗(x,x,x)

If possible suppose there existx,y∈X,x6=yand f(x) =g(x) =h(x) =xand f(y) =g(y) =

h(y) =y.

thenD∗(x,y,y) =D∗(f(x),g f(y),hg f(y)) ≤αD∗(x,y,y)

This impliesα≥1,which is contradiction, henceh,gand f have a unique fixed point.

Now we will prove the next theorem

Theorem 5. Let X be a complete metric space D∗-metric space f1,f2,f3:X →X be any three maps satisfy the following conditions

1)D∗ (f1(x),f2(y),f3(z)) ≤ α{D∗(x,y,z) +D∗(x,f1(x),f2(y)) +D∗(y,f2(y),f3(z)) For all

x,y,z∈X and 0≤α<1/3

2) f1,f2,f3are orbitally contractive.

3) D∗(f1(x),f1(x),f2(y))≤D∗(f1(x),f2(y),f3(z)),

D∗(f1(x),f1(x),f3(z))≤D∗(f1(x),f2(y),f3(z))

and D∗(f2(y),f2(y),f3(z))≤D∗(f1(x),f2(y),f3(z))for all x,y,z∈X

then f1,f2and f3have a unique common fixed point

Proof. Since f1,f2,f3 are orbitally contractive, there exists x0 ∈X, and a three positive real numbersλ1,λ2andλ3such thatD∗(a,b,fi(x))≤λiD∗(a,b,x),for allx∈O(f1,f2,f3,x0)and {a,b,x} 6={x0}. Defined a sequencexnby,

xn=

    

   

f1(xn−1),n=1,4, . . . .3k−2

f2(xn−1),n=2,5, . . . .3k−1

f3(xn−1),n=3,6, . . . .3k

    

   

To prove that{xn}is Cauchy sequence, assumem≥n.

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D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1)) ≤D∗(f1(xn−1),f2(xm−1),f3(xn−1))

≤α{D∗(xn−1,xn−1,xm−1) +D∗(xn−1,f1(xn−1),f2(xm−1))

+D∗(xm−1,f2(xm−1),f3(xn−1))}

≤α{D∗(xn−1,xn−1,xm−1) +λ1D∗(xn−1,xn−1,f2(xm−1))

+λ2D∗(xm−1,xm−1,f3(xn−1))}

≤α{D∗(xn−1,xn−1,xm−1) +λ1λ2D∗(xn−1,xn−1,xm−1)

+λ2λ3D∗(xm−1,xm−1,xn−1)}

=βD∗(xn−1,xn−1,xm−1),wereβ =α{1+λ1λ2+λ2λ3}<1. Continuo this way, we obtain,

D∗(xn,xn,xm)≤β D∗(xn−1,xn−1,xm−1)≤β2D∗(xn−2,xn−2,xm−2)

≤βnD∗(x0,x0,xm−n)≤βn(λ3)m−nD∗(x0,x0,f(x0))→0 asn,m→∞ Case2:xn= f1(xn−1),xm= f3(xm−1).

D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f3(xm−1)) ≤D∗(f1(xn−1),f2(xn−1),f3(xm−1))

≤α{D∗(xn−1,xn−1,xm−1) +D∗(xn−1,f1(xn−1),f2(xn−1))

+D∗(xn−1,f2(xn−1),f3(xm−1))}

≤α{D∗(xn−1,xn−1,xm−1) +λ1D∗(xn−1,xn−1,f2(xn−1))

+λ2D∗(xm−1,xn−1,f3(xm−1))}

≤α{D∗(xn−1,xn−1,xm−1) +λ1λ2D∗(xn−1,xn−1,xn−1)

+λ2λ3D∗(xm−1,xn−1,xm−1)}

=βD∗(xn−1,xn−1,xm−1),wereβ =α{1+λ1λ2+λ2λ3}<1. Continuo this way, as in case1, we obtain,

D∗(xn,xn,xm)≤βn(λ3)m−nD∗(x0,x0,f(x0))→0 asn,m→∞ Case3:xn= f2(xn−1),xm= f3(xm−1)is similar to the other cases. from the above cases,xconverges to some pointx∈X.

Now we want to prove thatxis fixed point for f1.

D∗(f1(x),x,x) =limn→∞D∗(f1(x),x(3n−1),x3n)

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≤limn→∞α{D∗(x,x3n−2,x3n−1) +D∗(x,f1(x),f2(x3n−2))

+D∗(x(3n−2),f2(x3n−2),f3(x3n−1)}

≤limn→∞α{D∗(x,x3n−2,x3n−1) +D∗(x,f1(x),x3n−1)

+D∗(x3n−2,x3n−1,x3n} ≤α D∗(x,f1(x),x)

This impliesD∗(x,f1(x),x) =0.Soxis fixed point for f1,similarly for f2,f3.

Now we want to prove thatxis a unique common fixed point of f1,f2,f3. Lety6=xbe such that

f1(x) = f2(x) = f3(x) =xand f1(y) = f2(y) = f3(y) =y,

thenD∗(x,y,y) =D∗(f1(x),f2(y),f3(y))≤α(D∗(x,y,y) +D∗(x,x,y) +D∗(y,y,y))

=2α{D∗(x,y,y)},which is a contradiction, unlessD∗(x,y,y) =0,which implies that

f1,f2,f3have a unique common fixed point.

Using the same procedures we can prove the following.

Theorem 6. Let X be a complete D∗−metric space and f1,f2,f3:X →X be any three maps satisfy the following conditions

1)D∗(f1(x),f2(y),f3(z))≤α1D∗(x,y,z) +α2{D∗(x,f1(x),f2(y)) +D∗(y,f2(y),f3(z))}

+α3{D∗(x,y,f2(y)) +D∗(y,z,f3(z))}

for all x,y,z∈X and0≤α1+2α2+2α3<1

2) f1,f2,f3are orbitally contractive .

3) D∗(f1(x),f1(x),f2(y))≤D∗(f1(x),f2(y),f3(z)),

D∗(f1(x),f1(x),f3(z))≤D∗(f1(x),f2(y),f3(z))

and D∗(f2(y),f2(y),f3(z))≤D∗(f1(x),f2(y),f3(z))for all x,y,z∈X then f1,f2and f3have a unique common fixed point

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xn=     

   

f1(xn−1),n=1,4, . . . .3k−2

f2(xn−1),n=2,5, . . . .3k−1

f3(xn−1),n=3,6, . . . .3k     

   

To prove that{xn} is Cauchy sequence supposem≥n,as in the prove of Theorem(5),we

shall prove the case,

xn= f1(xn−1),xm= f2(xm−1).Now,D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1) =D∗(f1(xn−1),f2(xm−1),f3(xn−1))

≤α1D∗(xn−1,xm−1,xn−1)+α2[D∗(xn−1,f1(xn−1),f2(xm−1)) +D∗(xm−1,f2(xm−1),f3(xn−1))]

+α3[D∗(xn−1,xm−1,f2(xm−1)) +D∗(xm−1,xn−1,f3(xn−1))]

≤α1D∗(xn−1,xm−1,xn−1)+α2[λ1D∗(xn−1,xn−1,f2(xm−1)) +λ2D∗(xm−1,xm−1,f3(xn−1))]

+α3[λ2D∗(xn−1,xm−1,xm−1) +λ3D∗(xm−1,xn−1,xn−1)}] ≤α1D∗(xn−1,xm−1,xn−1)

+α2[λ1λ2D∗(xn−1,xn−1,xm−1) +λ2λ3D∗(xm−1,xm−1,xn−1)] +α3[λ2D∗(xn−1,xm−1,xm−1) +λ3D∗(xm−1,xn−1,xn−1)}] ≤(α1+α2(λ1λ2+λ2λ3) +α3(λ2+λ3))D∗(xn−1,xm−1,xn−1) ≤β D∗(xn−1,xm−1,xn−1)whereβ=α1+2α2+2α3and 0≤β<1.

Continuo this way we obtained,

D∗(xn,xn,xm) ≤β D∗(xn−2,xm−2,xn−2)≤...

≤βnD∗(x0,xm−n,x0) =βn(max{λ1,λ2,λ3})m−n−1D∗(x0,x1,x0)→0 as

n,m→∞.

So{xn}is Cauchy hence, converges to some point sayx∈X.

Now we want to prove thatxis fixed point for f1,

D∗(f1(x),x,x) =limn→∞D∗(f1(x),x3n−1,x3n)

=limnD∗(f1(x),f2(x3n−2),f3(x3n−1)) ≤limn→∞α1D∗(x,x3n−2,x3n−1)

+α2[D∗(x,f1(x),f2(x3n−2)) +D∗(x3n−2,f2(x3n−2),f3(x3n−1))] +α3[D∗(x,x3n−2,f2(x3n−2)) +D∗(x3n−2,x3n−1,f3(x3n−1))]

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+α2[D∗(x,f1(x),x3n−1)) +D∗(x3n−2,x3n−1,x3n)]

+α3[D∗(x,x3n−2,x3n−1) +D∗(x3n−2,x3n−1,x3n)])

=α2D∗(x,f1(x),x).

ThereforeD∗(x,f1(x),x) =0, Soxfixed point for f1.

Similarly for f2,f3

For the uniqueness, suppose there exist y 6=x, such that x,y are common fixed points for

f1,f2,f3,then

D∗(x,y,y) =D∗( f1(x),f2(y),f3(y))

≤α1D∗(x,y,y) +α2{D∗(x,f1(x),f2(y))

+D∗(y,f2(y),f3(y))}+α3{D∗(x,y,f2(y)) +D∗(y,y,f3(y))}

= (α1+α2+α3)D∗(x,y,y)<D∗(x,y,y)

Theorem 7. let X be a complete D∗−metric space and f1,f2,f3:X →X be any three maps satisfies the followings

1)D∗(f1(x),f2(y),f3(z))

≤α max{D∗(x,y,z),D∗(x,f1(x),f2(y)),D∗(y,f2(y),f3(z)),D∗(x,y,f2(y)),D∗(y,z,f3(z))}

for all x,y,z∈X ,0≤α<1

2) f1,f2,f3are orbitally contractive orbitally.

3) D∗(f1(x),f1(x),f2(y))≤D∗(f1(x),f2(y),f3(z)),

D∗(f1(x),f1(x),f3(z))≤D∗(f1(x),f2(y),f3(z))

and D∗(f2(y),f2(y),f3(z))≤D∗(f1(x),f2(y),f3(z))for all x,y,z∈X then f1,f2and f3have a unique common fixed point

Proof. Since f1,f2,f3are orbitally contractive, there existsx0 ∈X,and a three positive real numbersλ1,λ2andλ3such thatD∗(a,b,fi(x))≤λiD∗(a,b,x),for allx∈O(f1,f2,f3,x0)and {a,b,x} 6={x0}. Defined a sequencexninX as,

xn=     

   

f1(xn−1),n=1,4, . . . .3k−2

f2(xn−1),n=2,5, . . . .3k−1

f3(xn−1),n=3,6, . . . .3k     

   

To prove that{xn} is Cauchy sequence supposem≥n,as in the prove of Theorem(5),we

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xn= f1(xn−1),xm= f2(xm−1).Now, with no loss of generality supposem≥n

D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1) ≤D∗(f1(xn−1),f2(xm−1),f3(xn−1))

≤αmax{D∗(xn−1,xm−1,xn−1),D∗(xn−1,f1(xn−1),f2(xm−1)),

D∗(xm−1,f2(xm−1),f3(xn−1)),D∗(xn−1,xm−1,f2(xm−1)),D∗(xm−1,xn−1,f3(xn−1))} ≤αmax{D∗(xn−1,xm−1,xn−1),λ1D∗(xn−1,xn−1,f2(xm−1)),λ2D∗(xm−1,xm−1,f3(xn−1)),

D∗(xn−1,xm−1,f2(xm−1)),D∗(xm−1,xn−1,f3(xn−1))}

≤α max{D∗(xn−1,xm−1,xn−1),λ1λ2D∗(xn−1,xn−1,xm−1),λ2λ3D∗(xm−1,xm−1,xn−1),

λ2D∗(xn−1,xm−1,xm−1),λ3D∗(xm−1,xn−1,xn−1)} ≤αD∗(xn−1,xm−1,xn−1)

continuo this way we obtained,

D∗(xn,xn,xm)≤α2D∗(xn−2,xm−2,xn−2)≤.... ≤αnD∗(x0,xm−n,x0)

=αn(max{λ1,λ2,λ3})m−n−1D∗(x0,x1,x0)→0as n,m→∞. So{xn}is Cauchy hence, converges to some point sayx∈X.

Now we want to prove thatxis fixed point for f1,

Suppose that f1(x)6=x

D∗(f1(x),x,x) =limn→∞D

(f

1(x),x3n−1,x3n)

=limn→∞D∗(f1(x),f2(x3n−2),f3(x3n−1))

≤limn→∞αmax{D

(x,x

3n−2,x3n−1),D∗(x,f1(x),f2(x3n−2)),D∗(x3n−2,f2(x3n−2),f3(x3n−1)),

D∗(xn−1,xm−1,f2(x3n−2)),D∗(x3n−1,x3n−1,f3(x3n−1))} ≤maxlimn→∞α{D∗(x,x3n−2,x3n−1),D∗(x,f1(x),f2(x3n−2)),

D∗(x3n−2,f2(x3n−2),f3(x3n−1)),D∗(xn−1,xm−1,f2(x3n−2)),D∗(x3n−1,x3n−1,f3(x3n−1))} =maxlimn→∞α{D∗(x,x3n−2,x3n−1),D∗(x,f1(x),x3n−1),D∗(x3n−2,x3n−1,x3n),

D∗(xn−1,x3n−2,x3n−1),D∗(x3n−1,x3n−1,x3n−)}

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=αD∗(x,f1(x),x)

<D∗(x,f1(x),x)

Contradiction

ThereforeD∗(x,f1(x),x) =0, Soxfixed point for f1.

Similarly for f2,f3

For the uniqueness, suppose there exist y 6=x, such that x,y are common fixed points for

f1,f2,f3,then

thenD∗(x,y,y) =D∗(f1(x),f2(y),f3(y))

≤αmax{D∗(x,y,y),D∗(x,f1(x),f2(y)),D∗(y,f2(y),f3(y)),D∗(x,y,f2(y)),D∗(y,y,f3(y))}

≤α max{D∗(x,y,y),D∗(x,x,y),D∗(y,y,y),D∗(x,y,y),D∗(y,y,y)}

=αD∗(x,y,y)<D∗(x,y,y)

This is a contradiction

Hence f1,f2and f3have a unique common fixed point.

Theorem 8. let X be a complete D∗−metric space and f1,f2,f3:X →X be any three maps satisfies the followings

1)D∗(f1(x),f2(y),f3(z))≤α1D∗(x,y,z) +α2max{D∗(x,f1(x),f2(y)),D∗(y,f2(y),f3(z))}

for all x,y,z∈X ,0≤α1+2α2<1

2) f1,f2,f3are orbitally contractive.

3)D∗(f1(x),f1(x),f2(y))≤D∗(f1(x),f2(y),f3(z)),

D∗(f1(x),f1(x),f3(z))≤D∗(f1(x),f2(y),f3(z))

and D∗(f2(y),f2(y),f3(z))≤D∗(f1(x),f2(y),f3(z))for all x,y,z∈X.

then f1,f2, f3have a unique common fixed point.

Proof. Since f1,f2,f3 are orbitally contractive, there exists x0 ∈X, and a three positive real numbersλ1,λ2andλ3such thatD∗(a,b,fi(x))≤λiD∗(a,b,x),for allx∈O(f1,f2,f3,x0)and {a,b,x} 6={x0}. Defined a sequencexninX as,

xn=

   

   

f1(xn−1),n=1,4, . . . .3k−2

f2(xn−1),n=2,5, . . . .3k−1

f3(xn−1),n=3,6, . . . .3k

   

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To prove that{xn} is Cauchy sequence supposem≥n,as in the prove of Theorem(5),we

shall prove the case,

with no loss of generality supposem≥n

D∗(xn,xn,xm) =D∗(f1(xn−1),f1(xn−1),f2(xm−1) ≤D∗(f1(xn−1),f2(xm−1),f3(xn−1)) ≤α1D∗(xn−1,xm−1,xn−1)

+α2max{D∗(xn−1,f1(xn−1),f2(xm−1)),D∗(xm−1,f2(xm−1),f3(xn−1))} ≤α1D∗(xn−1,xm−1,xn−1)

+α2max{λ1D∗(xn−1,xn−1,f2(xm−1)),λ2D∗(xm−1,xm−1,f3(xn−1))} ≤α1D∗(xn−1,xm−1,xn−1)

+α2max{λ1λ2D∗(xn−1,xn−1,xm−1),λ2λ3D∗(xm−1,xm−1,xn−1))} ≤α1D∗(xn−1,xm−1,xn−1)

+α2{λ1λ2D∗(xn−1,xn−1,xm−1) +λ2λ3D∗(xm−1,xm−1,xn−1))} ≤α1D∗(xn−1,xm−1,xn−1) +2α2D∗(xn−1,xn−1,xm−1)

= (α1+2α2)D∗(xn−1,xm−1,xn−1) Letβ = (α1+2α2)thenβ ≤1

D∗(xn,xm,xn)≤β D∗(xn−1,xm−1,xn−1) continuo this way we obtained

≤β2D∗(xn−2,xm−2,xn−2)

.

.

≤βn((max{λ1,λ2,λ3})m−n−1D∗(x0,x1,x0)→0as n,m→∞.

Now we want to prove thatxis fixed point for f1

Suppose that f1(x)6=x

D∗(f1(x),x,x) =limnD∗(f1(x),x3n−1,x3n)

=limn→∞D∗(f1(x),f2(x3n−2),f3(x3n−1)) ≤limn{α1D∗(x,x3n−2,x3n−1)

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+α2max{D∗(x,f1(x),x3n−1),D∗(x3n−2,x3n−1,x3n)}}

≤limnα1D∗(x,x3n−2,x3n−1)

+α2max{limn→∞D∗(x,f1(x),x3n−1),limn→∞D∗(x3n−2,x3n−1,x3n)}

=α2D∗(x,f1(x),x)

<D∗(x,f1(x),x) Contradiction

ThereforeD∗(x,f1(x),x) =0, Soxfixed point for f1. Similarly for f2,f3

For the uniqueness, suppose there exist y 6=x, such that x,y are common fixed points for

f1,f2,f3,then

thenD∗(x,y,y) =D∗(f1(x),f2(y),f3(y))

≤α1D∗(x,y,y) +α2max{D∗(x,f1(x),f2(y)),D∗(y,f2(y),f3(y))} =α1D∗(x,y,y) +α2D∗(x,y,y),

= (α1+α2)D∗(x,y,y)

<D∗(x,y,y).

This is a contradiction

Hence f1,f2and f3have a unique common fixed point.

Conflict of Interests

The authors declare that there is no conflict of interests.

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References

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