Song Jiping

(College of Mathematics and Information Science,Leshan Normal University,Leshan 614000,China)

Abstract：The notion of F-contraction is generalized from metric spaces to complex valued metric spaces,and F-quasi-contraction is introduced in complex valued metric spaces.Every F-contraction is F-quasi-contraction,and the reverse of this state is not true.By using the method of successive approximation,some coincidence point results for two mappings satisfying a F-contractive condition are proved,and some fixed point results for a F-contraction are obtained which generalize Wardowski′s results.Several illustrative examples are also given to highlight the realized generalization.

Keywords：complex valued metric space,F-contraction,coincidence point

1 Introduction

It is well known that contraction mapping principle is a very popular tool in solving existence problems in mathematics and technical applications.Since 1922 with Banach contraction principle,it is largely studied and generalized by several authors,see References[1-8].Reference[9]introduced the notion of complex valued metric space,which is more general than metric space,and established some common fixed point theorems for mappings satisfying some kind of rational compression conditions.Subsequently,more fixed point and common fixed point results for mappings satisfying certain contractive conditions are obtained by several authors in complex valued metric spaces,see References[10-14].References[15-16]introduced the notions of F-contraction and F-weak contraction to generalize the Banach′s contraction,and obtained some fixed point theorems in metric spaces.

The aim of this article is to generalize the notions of F-contraction from metric spaces to complex valued metric spaces.

2 Preliminaries

The following statements hold:

(i)0≤d(x,y),and d(x,y)=0 if and only if x=y for all x,y∈X;

(ii)d(x,y)=d(y,x)for all x,y∈X;

(iii)d(x,y)≤d(x,z)+d(z,y)for all x,y,z∈X.

Then d is called a complex valued metric on X,and(X,d)a complex valued metric space.

It is easy to see that complex valued metric spaces are more general than metric spaces.We refer the readers to Reference[9]for the related concepts such as convergence and Cauchy sequence in the complex valued metric space.

Remark 2.1 Let(X,d)be a complex valued metric space,then

is a sub-basis for a Hausdorfftopology T on X,where B(x,c)={y∈X:d(x,y)?c}.

Following is the concept of c-distance[17]in complex valued metric spaces.

(q1)0≤q(x,y)for all x,y∈X;

(q2)q(x,y)≤q(x,z)+q(z,y)for all x,y,z∈X;

(q3)for each x∈X and n∈,the set of positive integers,if q(x,yn)≤u for some u=ux∈,then q(x,y)≤u whenever{yn}is a sequence in X which converges to y∈X;

The following properties are from the Lemma 2.12 in Reference[17]:

Lemma 2.1 Let(X,d)be a complex valued metric space and q be a c-distance on X.Let{xn}and{yn}be two sequences in X,{un}and{vn}be two sequences inconverging to 0,and x,y,z∈X.

(3)if q(xn,xm)≤unfor m>n,then{xn}is a Cauchy sequence in X;

(4)if q(y,xn)≤un,then{xn}is a Cauchy sequence in X.

A partially ordered complex valued metric space is a complex valued metric space(X,d)endowed with a partial ordering?,denoted it by(X,d,?).

Definition 2.3[18]A partially ordered complex valued metric space(X,d,?)is regular if the following condition holds:for every non-decreasing sequence{xn}in X convergent to some x∈X,we have xn?x for all n∈N,and for every non-increasing sequence{xn}in X convergent to some x∈X,we have xn?x for all n∈N.

Let F:(0,+∞)→R be a function satisfying the following conditions:

(F1)s

We denote with F the family of functions satisfying the conditions(F1)-(F3).

Definition 2.4[15]Let(X,d)be a metric space.A mapping f:X→X is said to be a F-contraction if there exists a real number τ>0 and a function F ∈ F such that,for all x,y∈X with d(fx,fy)>0,we have

τ+F(d(fx,fy))≤F(d(x,y)).

Definition 2.5[16]Let(X,d)be a metric space.A mapping f:X→X is said to be a F-weak contraction if there exists a real number τ>0 and a function F ∈ F such that,for all x,y∈X with d(fx,fy)>0,we have

3 Main Results

(CF3)there exists a number α∈(0,1)such that

We denote with CF the family of all functions satisfying conditions(CF1)-(CF3).

Let α =max{α1,α2}.It is easy to see that

Then(X,d)is a complete complex valued metric space(see Reference[19]).f:X→X is given by

for x>0 or y>0.We see that f is an F-contraction.

Remark 3.1 It is easy to see that every F-contraction is a continuous mapping.In fact,from(CF1)and(1),we have d(fx,fy)≤d(x,y)for all x,y∈X with d(fx,fy)>0.If xn→x0as n→∞,for each c with 0?c in,there exists n0∈N such that d(xn,x0)?c for all n>n0.Then d(fxn,fx0)?c for all n>n0,that is fxn→fx0as n→∞.

Remark 3.2 Every F-contraction is F-quasi-contraction,the inverse does not hold.

Example 3.3 Let X=[0,+∞),and d(x,y)=|x?y|+2|x?y|i,then(X,d)is a complete complex valued metric space.f:X→X is defined by

and

Note that

we have

We obtain that(2)holds for 0≤x<1 and y≥1.Similarly,(2)holds for 0≤y<1 and x≥1.Hence,f is an F-quasi-contraction.

holds.If one of the following conditions is satisfied:

(1)g(X)is complete;

(2)(X,d)is complete,f and g is commuting mappings and,g is continuous.

Then f and g have a coincidence point x?∈ X with the unique point of coincidence gx?.

Proof Let x0∈X be arbitrary and fixed.We define a sequence{xn}in X by

gxn+1=fxn,n=0,1,2,···.

From(4),we have,for all n∈N,

for all n>n0.Hence,F(d(gxn+1,gxn)) ≤F(d(gx1,gx0))? nτ〈e for all n〉n0.From(CF2),we get

From(6)and(CF3),there exists α∈(0,1)such that

and

From(5),we deduce that

and

Taking limit as n→∞in the above inequality,we get

There exists a natural number n1such that

for all n>n1,that is

for all n>n1.Thus,

for all n>n1.For all n>n1and any natural number p,by using triangle inequality,we obtain that

it is a contradict.This completes the proof.

d(x,y)=|x?y|+|x?y|i,

then(X,d)is a complex valued metric space.f,g:X→X are given by

It is easy to see that,for all x,y∈X with 0

Corollary 3.1 Let(X,d)be a complete complex valued metric space,and f:X→X be an F-contraction,then f has an unique fixed point.

Proof By taking g=IX,the identity mapping on X,in Theorem 3.1,we see that the corollary is true.

(1)g(X)is complete and f(X)?g(X);

(2)f is g-monotone nondecreasing;

(3)there exists a x0∈X such that gx0?fx0;

(4)(X,d,?)is regular.

Then f and g have a coincidence point x?∈ X.

Proof Let x0∈X such that gx0?fx0.Since f(X)?g(X),there exists a x1∈X such that gx1=fx0,then gx0?gx1.From the condition(2),we have fx0?fx1.Similarly,there exists a x2∈X such that gx2=fx1and gx1?gx2.Continuing this process,we can construct a sequence{xn}such that gxn+1=fxnand gxn?gxn+1.

If there exists natural number n0such that gxn0+1=gxn0,then fxn0=gxn0,xn0is a coincidence point of f and g.So we assume that gxn+1gxnfor all n ∈.Since d(fxn,fxn?1)>0 and gxn?1? gxn,from(9),we have

If u(xn,xn?1)=d(gxn,gxn?1),then

The property(CF1)of F gives us that d(gxn+1,gxn)

From(CF1),we get

it deduces that d(gxn+1,gxn)

By using the triangle inequality,we have

We also obtain that

Since gxngxn+1and gxn? gxn+1for n ∈,we can suppose that gxnfx?for n ∈.Since(X,d,?)is regular,gxn? gx?.From(9),we have

From the property(CF1),we get

It deduces that

We have

Taking the limit as i→∞,we obtain that

Hence,gx?=fx?.Therefore,x?is a coincidence point of f and g.

(1)f is monotone nondecreasing;

(2)there exists a x0∈X such that x0?fx0;

(3)(X,d,?)is regular.

Then f has a fixed point.

Proof By taking g=IX,the identity mapping on X,in Theorem 3.2,we complete the proof.

(1)g(X)is complete and f(X)?g(X);

(2)f is g-monotone nondecreasing;

(3)there exists a x0∈X such that gx0?fx0;

(4)(X,d,?)is regular.

Then f and g have a coincidence point x?∈ X.If gv=fv,then q(gv,fv)=0.

Proof As in Theorem 3.2,we construct a sequence{xn}in X such that

holds,where

If u(xn?1,xn)=q(gxn?1,gxn),then

This implies that q(gxn?1,gxn)0.If,then

By the property(q2),q(gxn?1,gxn+1)≤q(gxn?1,gxn)+q(gxn,gxn+1).Consequently,

It is clear that(15)holds and q(gxn?1,gxn)0.

The above arguments give us q(gxn,gxn+1)0 for all.From(15),we have

nτ+F(q(gxn,gxn+1))≤F(q(gx0,gx1))

for n>n0,which implies that

For m>n>n0,since

we have

for n>n0.

Thus,q(gxni+1,fx?)

Thus,

By Lemma 2.1(1),fx?=gx?.

and so,

By Lemma 2.1(1),fx?=gx?.

Therefore,x?is a coincidence point of f and g.

Thus,q(gv,fv)=0.This completes the proof.

(1)f is monotone nondecreasing;

(2)there exists a x0∈X such that x0?fx0;

(3)(X,d,?)is regular.

Then f have a fixed point x?∈ X.If v=fv,then q(v,fv)=0.

Proof By taking g=IX,the identity mapping on X,in Theorem 3.2,we see that the corollary is true.