# Existence of Fixed Point Results inG-Metric Spaces

Authors: Zead Mustafa; Wasfi Shatanawi; Malik Bataineh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

The purpose of this paper is to prove the existence of fixed points
of contractive mapping defined onG-metric space where the completeness is
replaced with weaker conditions. Moreover, we showed that these conditions
do not guarantee the completeness of G-metric spaces.


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## Body

## 1. Introduction

The study of metric fixed point theory has been researched extensively in the past decades, since fixed point theory plays a major role in mathematics and applied sciences, such as optimization, mathematical models, and economic theories.Different mathematicians tried to generalize the usual notion of metric space(X,d) such as Gähler [1, 2] and Dhage [3–5] to extend known metric space theorems in more general setting, but different authors proved that these attempts are unvalid (for detail see [6–8]).In 2005, Mustafa and Sims introduced a new structure of generalized metric spaces (see [9]), which are called G-metric spaces as generalization of metric space (X,d), to develop and introduce a new fixed point theory for  various mappings in this new structure. The G-metric space is as follows.Definition 1.1 (see [9]).
LetX be a nonempty set, and let G:X×X×X→ℝ+, be a function satisfying the following: (G1)
G(x,y,z)=0 if  x=y=z,(G2)
0<G(x,x,y); for all x,y∈X,  with  x≠y,(G3)
G(x,x,y)≤G(x,y,z), for all  x,y,z∈X with z≠y,(G4)
G(x,y,z)=G(x,z,y)=G(y,z,x)=⋯,  (symmetry in all three variables),(G5)
G(x,y,z)≤G(x,a,a)+G(a,y,z), for all  x,y,z,a∈X,  (rectangle inequality).
Then the functionG is called a generalized metric,  or, more specifically a G-metric  on  X, and the pair (X,G) is a G-metric space.
Clearly these properties are satisfied whenG(x,y,z) is the perimeter of the triangle with vertices at x, y, and z in ℝ2; moreover taking a in the interior of the triangle shows that (G5) is the best possible.
If(X,d) is an ordinary metric space, then (X,d) can define G-metrics on X by
(Es)
Gs(d)(x,y,z)=d(x,y)+d(y,z)+d(x,z),(Em)
Gm(d)(x,y,z)=max{d(x,y),d(y,z),d(x,z)}.Proposition 1.2 (see [9]).
Let(X,G) be a G-metric space. Then for any x,y,z, and a∈X, it follows that(1)
ifG(x,y,z)=0, then x=y=z,(2)
G(x,y,z)≤G(x,x,y)+G(x,x,z),(3)
G(x,y,y)≤2G(y,x,x),(4)
G(x,y,z)≤G(x,a,z)+G(a,y,z),(5)
G(x,y,z)≤(2/3)(G(x,y,a)+G(x,a,z)+G(a,y,z)),(6)
G(x,y,z)≤(G(x,a,a)+G(y,a,a)+G(z,a,a)).Proposition 1.3 (see [9]).
EveryG-metric space (X,G) will define a metric space (X,dG) by
(1.1)dG(x,y)=G(x,y,y)+G(y,x,x),∀x,y∈X.Definition 1.4 (see [9]).
Let(X,G) be a G-metric space. Then for x0∈X, r>0, the G-ball with centerx0and radiusris
(1.2)BG(x0,r)={y∈X:G(x0,y,y)<r}.Proposition 1.5 (see [9]).
Let(X,G) be a G-metric space.  Then for any x0∈X and r>0,one has(1)
ifG(x0,x,y)<r, then x,y∈BG(x0,r),(2)
ify∈BG(x0,r), then there exists a  δ>0 such that BG(y,δ)⊆B(x0,r).Proof.
(1) follows directly from (G3), while (2) follows from (G5) with δ=r-G(x0,y,y).It follows from(2) of the above proposition that the family of all G-balls, ℬ={BG(x,r):x∈X,r>0}, is the base of a topology τ(G) on X, the  G-metric topology.Definition 1.6 (see [9]).
Let(X,G) be a G-metric space, let (xn) be sequence of points of X, a point x∈X is said to be the limit of the sequence (xn) if  limn,m→∞G(x,xn,xm)=0, and we say that the sequence (xn) is G-convergent to x.
Thus, ifxn→(G)0, in a G-metric space (X,G), then for any ϵ>0, there exists N∈ℕ such that G(x,xn,xm)<ϵ, for all n,m≥N, (through this paper we mean by ℕ the set of all natural numbers).Proposition 1.7 (see [9]).
Let(X,G) be a G-metric space. The sequence (xn)⊆X is G-convergent to  x if it converges to x in the G-metric topology, τ(G).Proposition 1.8 (see [9]).
Let(X,G) be a G-metric space. Then for a sequence (xn)⊆X and a point x∈X the following are equivalent(1)
(xn) is G-convergent to x,(2)
G(xn,xn,x)→0, as n→∞,(3)
G(xn,x,x)→0, as n→∞,(4)
G(xm,xn,x)→0, as m,n→∞.Definition 1.9 (see [9]).
Let(X,G) and (X′,G′) be G-metric spaces and let f:(X,G)→(X′,G′) be a function, then f is said to be G-continuous at a pointa∈X if and only if, given ϵ>0, there exists δ>0 such that x,y∈X; and G(a,x,y)<δ implies G′(f(a),f(x),f(y))<ϵ. A function f is G-continuous at X if and only if it is G-continuous at all a∈X.Proposition 1.10 (see [9]).
Let(X,G), (X′,G′) be G-metric spaces. Then a function f:X→X′ is G-continuous at a point x∈X if and only if it is G-sequentially continuous at x; that is, whenever (xn) is G-convergent to xone has(f(xn)) is G-convergent to f(x).Proposition 1.11 (see [9]).
Let(X,G) be a G-metric space. Then the function G(x,y,z) is jointly continuous in all three of its variables.Definition 1.12 (see [9]).
Let(X,G) be a G-metric space. Then the sequence (xn)⊆X is said to be G-Cauchy  if for every ϵ>0, there  exists N∈ℕ  such that G(xn,xm,xl)<ϵ for all n,m,l≥N.Definition 1.13 (see [9]).
AG-metric space (X,G) is said to be G-complete  (or complete G-metric space) if every G-Cauchy sequence in (X,G) is G-convergent in (X,G).


## 2. The Main Results

In this section we will prove several theorems in each of which we have omitted the completeness property ofG-metric space and we have obtained the same conclusion as in complete G-metric space, but with assumed sufficient conditions.Theorem 2.1.
Let(X,G) be a G-metric space and let T:X→X be a mapping such that T satisfies that(A1)
G(Tx,Ty,Tz)≤aG(x,Tx,Tx)+bG(y,Ty,Ty)+cG(z,Tz,Tz) for all x,y,z∈X where 0<a+b+c<1,(A2)
T is G-continuous at a point u∈X,(A3)
there isx∈X;{Tn(x)} has a subsequence {Tni(x)}G-converges to u. Then u is a unique fixed point  (i.e., Tu=u).Proof.
G-continuity of T at u implies that {Tni+1(x)}G-convergent to T(u). Suppose T(u)≠u, consider the two G-open balls B1=B(u,ϵ) and B2=B(Tu,ϵ) where ϵ<(1/6)min{G(u,Tu,Tu),G(Tu,u,u)}.
SinceTni(x)→u and Tni+1(x)→Tu, then there exist N1∈ℕ such that if i>N1 implies Tni(x)∈B1 and Tni+1(x)∈B2. Hence our assumption implies that we must have
(2.1)G(Tni(x),Tni+1(x),Tni+1(x))>ϵ,∀i>N1.
On the other hand we have from (A1),(2.2)G(Tni+1(x),Tni+2(x),Tni+3(x))≤aG(Tni(x),Tni+1(x),Tni+1(x))+bG(Tni+1(x),Tni+2(x),Tni+2(x))+cG(Tni+2(x),Tni+3(x),Tni+3(x))
but, by axioms of G-metric (G3), we have
(2.3)G(Tni+1(x),Tni+2(x),Tni+2(x))≤G(Tni+1(x),Tni+2(x),Tni+3(x)),(2.4)G(Tni+2(x),Tni+2(x),Tni+3(x))≤G(Tni+1(x),Tni+2(x),Tni+3(x)).
So, from (2.3) and (2.4), we see (2.2) becomes
(2.5)G(Tni+1(x),Tni+2(x),Tni+3(x))≤qG(Tni(x),Tni+1(x),Tni+1(x)),
where q=a/(1-(b+c)) and q<1, since 0<a+b+c<1.
Hence (2.3) and (2.5) implies that
(2.6)G(Tni+1(x),Tni+2(x),Tni+2(x))≤qG(Tni(x),Tni+1(x),Tni+1(x)).
Forl>j>N1 and by repeated application of (2.6) we have
(2.7)G(Tnl(x),Tnl+1(x),Tnl+1(x))≤qG(Tnl-1(x),Tnl(x),Tnl(x))≤q2G(Tnl-2(x),Tnl-1(x),Tnl-1(x))≤⋯≤qnl-njG(Tnj(x),Tnj+1(x),Tnj+1(x)).
So, as l→∞ we have limG(Tnl(x),Tnl+1(x),Tnl+1(x))≤0 which contradict (2.1), hence Tu=u.
Suppose there isv∈X;Tv=v, then from (A1), we have
(2.8)G(u,v,v)=G(Tu,Tv,Tv)≤aG(u,Tu,Tu)+(b+c)G(v,Tv,Tv)=0.
This prove the uniqueness of u.In [10] we have proved the following theorem.Theorem 2.2 (see [10]).
Let(X,G) be a complete G-metric space and let T:X→X be a mapping satisfies the following condition for all x,y,z∈X:
(2.9)G(T(x),T(y),T(z))≤aG(x,T(x),T(x))+bG(y,T(y),T(y))+cG(z,T(z),T(z))+dG(x,y,z),
where 0≤a+b+c+d<1, then T has a unique fixed point, say u, and T is G-continuous at u.We see that if we taked=0, the following theorem becomes a direct result.Theorem 2.3.
Let(X,G) be complete G-metric space and let T:X→X be a mapping satisfies for all x,y,z∈X(2.10)G(Tx,Ty,Tz)≤aG(x,Tx,Tx)+bG(y,Ty,Ty)+cG(z,Tz,Tz),
where 0<a+b+c<1, then T has a unique fixed point, say u, and T is G-continuous at u.If we compare Theorem2.3 with Theorem 2.1, we see that in Theorem 2.1 we have omitted the completeness property of the G-metric space and instead we have assumed conditions (2) and (3). However, the following examples support that conditions (2) and (3) in Theorem 2.1 do not guarantee the completeness of the G-metric space.Example 2.4.
LetX=[0,1),T(x)=x/4 and G(x,y,z)=max{|x-y|,|y-z|,|x-z|}. Then (X,G) is G-metric space but not complete, since the sequence xn=1-1/n is G-cauchy which is not G-convergent in (X,G). However, conditions (2) and (3) in Theorem 2.1 are satisfied.Theorem 2.5.
Let(X,G) be a G-metric space and let T:X→X be a G-continuous mapping satisfies the following conditions:(B1)
G(Tx,Ty,Tz)≤k{G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz) for all x,y,z∈M, where M is an every where dense subset of X (with respect the topology of G-metric convergence) and 0<k<1/6,(B2)
there isx∈X such that {Tn(x)}→xo. Then xo is unique fixed point.Proof.
The proof will follow from Theorem2.1, if we can show that condition (A1) in Theorem 2.1 holds for any x,y,z∈X.
Letx,y,z be any elements of X.
Case 1.
Ifx,y,z∈X∖M, let {xn}, {yn}, and {zn} be a sequences in M such that yn→y,  xn→x and zn→z. From (G5) we have
(2.11)G(Tx,Ty,Tz)≤G(Tx,Ty,Ty)+G(Tz,Ty,Ty),
also
(2.12)G(Tz,Ty,Ty)≤G(Tz,Tzn,Tzn)+G(Tzn,Tyn,Tyn)+G(Tyn,Ty,Ty)
and by (B1), we have
(2.13)G(Tzn,Tyn,Tyn)≤k{G(zn,Tzn,Tzn)+2G(yn,Tyn,Tyn)},
again by (G5) we have
(2.14)G(zn,Tzn,Tzn)≤G(zn,z,z)+G(z,Tz,Tz)+G(Tz,Tzn,Tzn),G(yn,Tyn,Tyn)≤G(yn,y,y)+G(y,Ty,Ty)+G(Ty,Tyn,Tyn).
So, from (2.13) and (2.14) we see that (2.12) becomes
(2.15)G(Tz,Ty,Ty)≤(1+k)G(Tz,Tzn,Tzn)+G(Tyn,Ty,Ty)+kG(zn,z,z)+2kG(yn,y,y)+2kG(Ty,Tyn,Tyn)+kG(z,Tz,Tz)+2kG(y,Ty,Ty),
by the same argument we deduce that
(2.16)G(Tx,Ty,Ty)≤(1+k)G(Tx,Txn,Txn)+G(Tyn,Ty,Ty)+kG(xn,x,x)+2kG(yn,y,y)+2kG(Ty,Tyn,Tyn)+kG(x,Tx,Tx)+2kG(y,Ty,Ty).
Hence, by (2.15) and (2.16), we have
(2.17)G(Tx,Ty,Tz)≤G(Tx,Ty,Ty)+G(Tz,Ty,Ty)≤{(1+k)G(Tx,Txn,Txn)+G(Tyn,Ty,Ty)+kG(xn,x,x)+2kG(yn,y,y)+2kG(Ty,Tyn,Tyn)+kG(x,Tx,Tx)+2kG(y,Ty,Ty)}+{(1+k)G(Tz,Tzn,Tzn)+G(Tyn,Ty,Ty)+kG(zn,z,z)+2kG(yn,y,y)+2kG(Ty,Tyn,Tyn)+kG(z,Tz,Tz)+2kG(y,Ty,Ty)}.
Now lettingn→∞ in the above inequality and using the fact that T is G-continuous we get
(2.18)G(Tx,Ty,Tz)≤k{G(x,Tx,Tx)+4G(y,Ty,Ty)+G(z,Tz,Tz)}.
Case 2.
Ifx,y∈M and z∈X∖M, let {zn} be a sequence in M such that zn→z, then by (G5), we have
(2.19)G(Tx,Ty,Tz)≤G(Tx,Ty,Ty)+G(Tz,Ty,Ty)
but by (B1), we have
(2.20)G(Tx,Ty,Ty)≤k{G(x,Tx,Tx)+2G(y,Ty,Ty)},
and by (G5), we have
(2.21)G(Tz,Ty,Ty)≤G(Tz,Tzn,Tzn)+G(Tzn,Ty,Ty).
Again by (B1), we have
(2.22)G(Tzn,Ty,Ty)≤k{G(zn,Tzn,Tzn)+2G(y,Ty,Ty)}.
Also, by (G5), we have
(2.23)G(zn,Tzn,Tzn)≤G(zn,z,z)+G(z,Tz,Tz)+G(Tz,Tzn,Tzn).
So, from (2.21), (2.22), and (2.23), we see that (2.19) becomes
(2.24)G(Tx,Ty,Tz)≤k{G(x,Tx,Tx)+2G(y,Ty,Ty)+G(zn,z,z)+G(z,Tz,Tz)+G(Tz,Tzn,Tzn)}+G(Tz,Tzn,Tzn)+2kG(y,Ty,Ty).
Now lettingn→∞ in the above inequality, we get
(2.25)G(Tx,Ty,Tz)≤k{G(x,Tx,Tx)+4G(y,Ty,Ty)+G(z,Tz,Tz)}.
Case 3.
Ify∈M and x,z∈X∖M, let {xn} and {zn} be a sequences in M such that xn→x and zn→z, but by (G5), we have
(2.26)G(Tx,Ty,Tz)≤G(Tx,Ty,Ty)+G(Tz,Ty,Ty),(2.27)G(Tx,Ty,Ty)≤G(Tx,Txn,Txn)+G(Txn,Ty,Ty),
also, from (B1), we have
(2.28)G(Txn,Ty,Ty)≤k{G(xn,Txn,Txn)+2G(y,Ty,Ty)},
and from (G5), we have
(2.29)G(xn,Txn,Txn)≤G(xn,x,x)+G(x,Tx,Tx)+G(Tx,Txn,Txn).
So, by (2.28) and (2.29), we have
(2.30)G(Txn,Ty,Ty)≤2kG(y,Ty,Ty)+kG(xn,x,x)+kG(x,Tx,Tx)+kG(Tx,Txn,Txn),
then from (2.27) and (2.30) we have
(2.31)G(Tx,Ty,Ty)≤kG(xn,x,x)+kG(x,Tx,Tx)+(1+k)G(Tx,Txn,Txn)+2kG(y,Ty,Ty).
By the same argument we deduce that
(2.32)G(Tz,Ty,Ty)≤kG(zn,z,z)+kG(z,Tz,Tz)+(1+k)G(Tz,Tzn,Tzn)+2kG(y,Ty,Ty).
Then, from (2.31) and (2.32), we see (2.26) becomes
(2.33)G(Tx,Ty,Tz)≤G(Tx,Ty,Ty)+G(Tz,Ty,Ty)≤kG(xn,x,x)+kG(x,Tx,Tx)+kG(Tx,Txn,Txn)+2kG(y,Ty,Ty)+kG(zn,z,z)+kG(z,Tz,Tz)+kG(Tz,Tzn,Tzn)+2kG(y,Ty,Ty).
Now lettingn→∞ in the above inequality and using the fact that T is G-continuous we get
(2.34)G(Tx,Ty,Tz)≤k{G(x,Tx,Tx)+4G(y,Ty,Ty)+G(z,Tz,Tz)}.
So, in all cases we have for allx,y,z∈X(2.35)G(Tx,Ty,Tz)≤aG(x,Tx,Tx)+bG(y,Ty,Ty)+cG(z,Tz,Tz),
where a=k,b=4k,c=k, and a+b+c<1 since 0<k<1/6, then by Theorem 2.1,  T has a unique fixed point.Corollary 2.6.
Let(X,G) be G-metric space and let T:X→X be a mapping such that T satisfies that(C1)
G(Tx,Ty,Ty)≤aG(x,Tx,Tx)+bG(y,Ty,Ty) for all x,y∈X where 0<a+b<1,(C2)
T is G-continuous at a point u∈X,(C3)
there isx∈X;{Tn(x)} has a subsequence {Tni(x)}G-converges to u. Then u is a unique fixed point.Proof.
Letz=y in condition (A1), then we see that every mapping satisfies condition (C1) will satisfy condition (A1), so the proof follows from Theorem 2.1.Corollary 2.7.
Let(X,G) be G-metric space and let  T:X→X be a G-continuous mapping satisfies that(D1)
G(Tx,Ty,Ty)≤k{G(x,Tx,Tx)+G(y,Ty,Ty)} for all  x,y∈M where M is an every where dense subset of X (with respect the topology of G-metric convergence) and 0<k<1/6,(D2)
there isx∈X such that {Tn(x)}→xo. Then xo is unique fixed point.Proof.
Letz=y in condition (B1), then we see that every mapping satisfies condition (D1) will satisfy condition (B1), so the proof follows from Theorem 2.5.Corollary 2.8.
Let(X,G) be G-metric space and let T:X→X be a mapping such that T satisfies that(E1)
G(Tx,Ty,Ty)≤kG(x,y,y) for all x,y∈X where 0<k<1/4,(E2)
T is G-continuous at a point u∈X,(E3)
there isx∈X;{Tn(x)} has a subsequence {Tni(x)}G-converges to u. Then u is a unique fixed point.Proof.
By axioms ofG-metric (G5), we have
(2.36)G(x,y,y)≤G(x,Tx,Tx)+G(Tx,Ty,Ty)+G(Ty,y,y),G(Ty,y,y)≤2G(y,Ty,Ty),
so, from (2.36), we see that (E1) becomes
(2.37)G(Tx,Ty,Ty)≤kG(x,y,y)≤kG(x,Tx,Tx)+kG(Tx,Ty,Ty)+2kG(y,Ty,Ty),
then T will satisfy the following condition
(2.38)G(Tx,Ty,Ty)≤aG(x,Tx,Tx)+bG(y,Ty,Ty)
for all x,y∈X, where a=k/(1-k),  b=2k/(1-k), and a+b<1, since k<1/4.
So, condition (C1) is satisfied and the proof follows from Corollary2.6.



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