GAO Zheng-ying,JI Pei-sheng

(School of Mathematics and Statistics,Qingdao University,Qingdao 266071,China)

Abstract:In this article,we study concepts of weak F-contraction of type(A)and weak F-contraction of type(B)in generalized metric spaces.By using the method of iteration,we obtain some fixed point theorems for these mappings in a complete generalized metric space,which generalize the results of F-contraction in complete metric spaces.

Keywords: generalized metric space; fixed point;F-contraction

1 Introduction and Preliminaries

It is well known that Banach contraction mapping principle[1]which was published in 1922 is one of the most important theorems in classical functional analysis.Indeed,it was widely used as the source of metric fixed point theory,for more details see[2–8].In 2012,Wardowski[2]introduced a new concept of contraction called F-contraction and proved a fixed point theorem which generalized the Banach contraction principle.Later,Wardowski and Van Dung[3]introduced the definition of an F-weak contraction and proved a fixed point theorem for it.Dung and Hung[4]generalized an F-weak contraction to a generalized F-contraction in 2015.Recently,Piri and Kumam[5,6]described a large class of functions by replacing condition(F3)′instead of condition(F3)in the definition of F-contraction.They introduced the concepts of modified generalized F-contraction of type(A)and modified generalized F-contraction of type(B)and proved some fixed point theorems for these contractions in a complete metric space.

Following this direction of research,in this paper,we introduce some F-contractions which satisfy different conditions in generalized metric space and then we prove some fixed point theorems for these contractions in a complete generalized metric space.Finally,we give an example to support our result.

Throughout the article N,R+and R will denote the set of natural numbers,non-negative real numbers and real numbers,respectively.

First,Wardowski[2]introduced the notion of a F-contraction and proved the Wardowskifixed point theorem as the generalization of Banach contraction principle.Let F:(0,∞)→R be a mapping satisfying

(F1)F is strictly increasing,that is,s0;

(F2)for every sequence{sn}in R+,we have=0 if and only if

(F3)there exists a number k∈(0,1)such that

(F3)′F is continuous on(0,∞).

We denote with F1the family of all functions F that satisfy conditions(F1)–(F3).Let z denote the family of all functions F:R+→R that satisfy conditions(F1),(F3)and F denote the family of all functions F:R+→ R that satisfy conditions(F1),(F3)′.

Example 1 The following functions F:(0,∞)→R belongs to F1:

(i)F(t)=lnt with t>0,

(ii)F(t)=lnt+t with t>0.

Definition 1[2]Let(X,d)be a metric space.A mapping T:X→X is called an F-contraction on X if there exist F ∈ F1and τ>0 such that for all x,y ∈ X with d(Tx,Ty)>0,we have τ+F(d(Tx,Ty))≤ F(d(x,y)).

Theorem 1 Let(X,d)be a complete metric space and T:X→X be an F-contraction.Then T has a unique fixed point x?∈ X and for every x ∈ X the sequence{Tnx}n∈Nconverges to x?.

In 2000,Branciari[7]introduced the concept of rectangular metric space by replacing the sum of the right hand side of the triangle inequality in metric space by a three-term expression and proved an analog of the Banach contraction principle in such spaces.

Definition 2[7]Let X be a nonempty set and d:X ×X → [0,∞)satisfying the following conditions,for all x,y∈X and all distinct u,v∈X each of which is different from x and y:

(GMS1)d(x,y)=0 if and only if x=y,

(GMS2)d(x,y)=d(y,x),

(GMS3)d(x,y)≤d(x,u)+d(u,v)+d(v,y).

Then the map d is called generalized metric and abbreviated as GM,here the pair(X,d)is called a generalized metric space and abbreviated as GMS.

Definition 3[7]Let(X,d)be a generalized metric space and{xn}be a sequence of elements of X.

(1)A sequence{xn}is said to be GMS convergent to a limit x∈X if and only if d(xn,x)→0 as n→∞.

(2)A sequence{xn}is said to be GMS Cauchy sequence if and only if for every ε>0 there exists a positive integer N(ε)such that d(xn,xm)< ε for all n>m>N(ε).

(3)A GMS(X,d)is called complete if every GMS Cauchy sequence in X is GMS convergent.

(4)A mapping T:(X,d)→(X,d)is continuous if for any sequence{xn}in X such that d(xn,x)→0 as n→∞,we have d(Txn,Tx)→0 as n→∞.

Lemma 1 Let(X,d)be a GMS and let{xn}be a sequence in X.Assume that{xn}is Cauchy sequence with xn≠xmfor all n,m∈N with n≠m andThenfor all y∈X.

2 Main Results

Definition 4 Let(X,d)be a GMS.A mapping T:X→X is called a weak F-contraction of type(A)on X if there exist F ∈ z and τ>0 such that

Theorem 2 Let(X,d)be a complete GMS and T:X→X be a weak F-contraction of type(A).Then T has a unique fixed point x?∈X and for every x∈X the sequence{Tnx}n∈Nconverges to x?.

ProofChoose x0∈ X and define a sequenceby

If there exists n∈N such that xn=xn+1.Then xnis a fixed point of T and we have nothing to prove.Now we suppose that xn≠xn+1,i.e.,d(Txn?1,Txn)>0 for all n ∈ N.It follows from(2.1)that,for all n∈N,

i.e.,

It follows from(2.2)and(F1)that

Therefore{d(xn,xn+1)}n∈Nis a nonnegative decreasing sequence of real numbers,and hence

Now,we claim that γ=0.Arguing by contradiction,we assume that γ >0.

Since{d(xn,xn+1)}n∈Nis a nonnegative decreasing sequence of real numbers,for every n∈N,we have

From(2.2),(2.3)and(F1),we get

for all n ∈ N.Since F(γ) ∈ R and= ?∞,there exists n1∈ N such that

It follows from(2.4)and(2.5)that F(γ)n1.It is a contraction.Therefore we have

From(F3),there exists k∈(0,1)such that

It follows from(2.4)that

for all n∈N.By using(2.6),(2.7)and taking the limit as n→∞in(2.8),we get

Then there exists n2∈N such that n(d(xn,xn+1))k≤1 for all n≥n2,that is,

Now,we shall prove that xn≠xm,for all n≠m.Assume on the contrary that xn=xmfor some m,n∈N with n≠m.Since d(xp,xp+1)>0 for each p∈N,without loss of generality,we may assume that m>n+1.Substitute again x=xn=xm,y=xn+1=xm+1in(2.1),which yield

then

Since τ>0,we get a contradiction,therefore xn≠xm,for all n ≠m.Now,we prove=0 for all n∈N.It follows from(2.1)that for all n∈N,

i.e.,

It follows from(2.11)and(F1)that

Therefore{d(xn,xn+2)}n∈Nis a nonnegative decreasing sequence of real numbers,and hence

Now,we claim that γ=0.Arguing by contradiction,we assume that γ >0.

Since{d(xn,xn+2)}n∈Nis a nonnegative decreasing sequence of real numbers,for every n∈N,we have

From(2.11),(2.12)and(F1),we get

for all n ∈ N.Since F(γ) ∈ R and= ?∞,there exists n3∈ N such that

It follows from(2.13)and(2.14)that F(γ)n3.It is a contraction.Therefore we have

Now,we shall prove that{xn}is a GMS Cauchy sequence in(X,d),that is,for ε>0 and N(ε)such that d(xn,xm)< ε,for all m,n ∈ N with m>n>N(ε).Assume that m=n+k,where k>2.We will consider the only two cases as follows.

Case 1 For k is odd.Let k=2l+1,where l∈N.By using the quadrilateral inequality(GMS3)and(2.10),we have

Case 2 For k is even.Let k=2l,where l∈N.By using the quadrilateral inequality(GMS3)and(2.10),(2.15),we have

Therefore we conclude that{xn}is a GMS Cauchy sequence in(X,d).By the completeness of(X,d),there exists x?∈ X such that

Also,we suppose that T is continuous,we have

By Lemma 1,it follows that xndiffers from both x?and Tx?for n sufficiently large.Hence,we have d(x?,Tx?)≤ d(x?,xn)+d(xn,xn+1)+d(xn+1,Tx?),from(2.6),(2.16),(2.17),we haved(x?,Tx?)≤=0.So we obtain that d(x?,Tx?)=0,that is Tx?=x?.Hence x?is a fixed point of T.Now let us show that T has at most one fixed point.Indeed,if x?,y?∈ X are two distinct fixed points of T,that is,Tx?=x?≠y?=Ty?,then d(Tx?,Ty?)>0.So from the assumption of the theorem,we obtain

which is a contradiction.Thus T has a unique fixed point.

Now,we show another important result.

Definition 5 Let(X,d)be a GMS.A mapping T:X→X is called a weak F-contraction of type(B)on X if there exist F ∈ F and τ>0 such that

Theorem 3 Let(X,d)be a complete GMS and T:X→X be a weak F-contraction of type(B),then T has a unique fixed point x?∈X and for every x∈X the sequence{Tnx}n∈Nconverges to x?.Proof Choose x0∈ X and define a sequenceby

By using the similar method in the proof of Theorem 2,we have

Now,we shall prove that{xn}is a GMS Cauchy sequence in(X,d).Suppose{xn}is not a Cauchy sequence,then there exists ε>0 for which we can find two subsequencesand{x}of{x}such that p(n)is the smallest index for whichq(n)n

Since xm≠xnfor all m,n∈N with n≠m,we have

By taking the limit in above inequality and using inequalities(2.19),we get

Since

we have

So we have

Finally,we claim that x?=Tx?,we only have the following two cases

Example 2 Let X be a finite set defined as X={1,2,3,4}.Define d:X×X →[0,∞)as

The function d is not a metric on X.Indeed,note that

that is,the triangle inequality is not satis fied.However d is a generalized metric on X,moreover,(X,d)is a complete generalized metric space.Define T:X→X as

For x∈{1,2,3}and y=4,we haved(Tx,Ty)=d(2,3)=1>0,d(x,y)=4.Therefore.So by choosing F(t)=ln(t)and,we see that T is a F-contraction which satis fies Theorems 3.