# On a Fixed Point for Generalized Contractions in Generalized Metric Spaces

Authors: Wasfi Shatanawi; Ahmed Al-Rawashdeh; Hassen Aydi; Hemant Kumar Nashine
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

Lakzian and Samet (2010)  studied some fixed-point results in generalized metric spaces in the sense of Branciari.  In this paper, we study the existence of fixed-point results of mappings satisfying generalized weak contractive conditions in the framework of a generalized metric space in sense of Branciari. Our results modify and generalize the results of Laksian and Samet, as well as, our results generalize several well-known comparable results in the literature.


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## Body

## 1. Introduction and Preliminaries

Branciari in [1] initiated the notion of a generalized metric space as a generalization of a metric space in such a way that the triangle inequality is replaced by the “quadrilateral inequality,” d(x,y)≤d(x,a)+d(a,b)+d(b,y) for all pairwise distinct points x,y,a, and b of X. Afterwards, many authors initiated and studied many existing fixed-point theorems in such spaces. For more details about fixed-point theory in generalized metric spaces, we refer the reader to [1–13].The following definitions will be needed in the sequel.Definition 1.1 (see [1]).
LetX be a nonempty set and d:X×X→[0,+∞) such that for all x,y∈X and for all distinct points u,v∈X each of them different from x and y, one has(p1):
x=y⇔d(x,y)=0,(p2):
d(x,y)=d(y,x),(p3):
d(x,y)≤d(x,u)+d(u,v)+d(v,y). Then, (X,d) is called a generalized metric space (or shortly g.m.s).
Any metric space is a generalized metric space, but the converse is not true [1].Definition 1.2 (see [1]).
Let(X,d) be a g.m.s, {xn} a sequence in X, and x∈X. We say that {xn} is g.m.s convergent to x if and only if d(xn,x)→0 as n→+∞. We denote this by xn→x.Definition 1.3 (see [1]).
Let(X,d) be a g.m.s and {xn} a sequence in X. We say that {xn} is a g.m.s Cauchy sequence if and only if for each ɛ>0 there exists a natural number N such that d(xn,xm)<ɛ for all n>m>N.Definition 1.4 (see [1]).
Let(X,d) be a g.m.s. Then, (X,d) is called a complete g.m.s if every g.m.s Cauchy sequence is g.m.s convergent in X.Very recently, Lakzian and Samet [9] proved the following nice result.Theorem 1.5.
Let(X,d) be a Hausdorff and complete generalized metric space. Suppose that T:X→X is such that for all x,y∈X(1.1)ψ(d(Tx,Ty))≤ψ(d(x,y))-ϕ(d(x,y)),
where ψ:[0,∞)→[0,∞) is  continuous  and  nondecreasing with ψ(t)=0 if and only if t=0, and ϕ:[0,∞)→[0,∞) is continuous and ϕ(t)=0 if and only if t=0. Then, there exists a unique point u∈X such that u=Tu.Note that Theorem1.5 extends a result of Dutta and Choudhury [14] to the set of generalized metric spaces. Moreover, its proof is more technical compared with that of [9].In this paper, we generalize in some cases Theorem1.5 by replacing in (1.1) the term d(x,y) by the quantity max{d(x,y),d(x,Tx),d(y,Ty)} and the continuity of ϕ by lower semicontinuity. Also, we derive some useful corollaries of this result.


## 2. Main Results

LetX be a nonempty set and T:X→X a given mapping. For all x,y∈X, set
(2.1)M(x,y)=max⁡{d(x,y),d(x,Tx),d(y,Ty)}.
Also, let  Ψ={ψ∣ψ:[0,∞)→[0,∞)becontinuous,nondecreasing,andψ(t)=0ifandonlyift=0}, and Φ={ϕ∣ϕ:[0,∞)→[0,∞)islowersemicontinuous,ϕ(t)>0forallt>0andϕ(0)=0}. Note that, if ψ∈Ψ, ψ is called an altering distance function [15].The notion of a periodic point of a given mappingT:X→X is crucial for proving our main theorem. So we need the following definition.Definition 2.1.
LetX be a nonempty set. A given mapping T:X→X admits a periodic point if there exists u∈X such that u=Tpu for some p≥1. If p=1, u is a fixed point.
Hence, each fixed point is also a periodic point ofT.Now, in the following, let us prove our main result.Theorem 2.2.
Let(X,d) be a Hausdorff and complete generalized metric space. Suppose that T:X→X is such that for all x,y∈X(2.2)ψ(d(Tx,Ty))≤ψ(M(x,y))-ϕ(M(x,y)),
where ψ∈Ψ, ϕ∈Φ, and M(x,y) is defined by (2.1). Then, there exists a unique point u∈X such that u=Tu.Proof.
First, it is obvious thatM(x,y)=0 if and only if x=y is a fixed point of T. Let x0∈X an arbitrary point. By induction, we easily construct a sequence {xn} such that
(2.3)xn+1=Txn=Tn+1x0∀n≥0.Step  1. We claim that
(2.4)lim⁡n→∞d(xn,xn+1)=0.
Substituting x=xn and y=xn-1 in (2.2) and using properties of functions ψ and ϕ, we obtain
(2.5)ψ(d(xn+1,xn))=ψ(d(Txn,Txn-1))≤ψ(M(xn,xn-1))-ϕ(M(xn,xn-1))≤ψ(M(xn,xn-1))
which implies that
(2.6)d(xn+1,xn)≤M(xn,xn-1)∀n≥1.
Note that
(2.7)M(xn,xn-1)=max⁡⁡{d(xn,xn-1),d(xn,Txn),d(xn-1,Txn-1)}=max⁡⁡{d(xn,xn-1),d(xn,xn+1)}.
If for some n≥1, d(xn-1,xn)<d(xn,xn+1), then M(xn,xn-1)=d(xn,xn+1)>0 and ϕ(d(xn+1,xn))>0 by a property of ϕ, so (2.5) becomes
(2.8)0<ψ(d(xn+1,xn))≤ψ(d(xn+1,xn))-ϕ(d(xn+1,xn+1))<ψ(d(xn+1,xn))
a contradiction. Thus, for all n≥1,
(2.9)d(xn+1,xn)≤d(xn-1,xn)=M(xn-1,xn).
From (2.9), the sequence {d(xn,xn+1)} is monotone nonincreasing and so bounded below. So there exists r≥0 such that
(2.10)lim⁡n→∞d(xn,xn+1)=lim⁡n→∞M(xn-1,xn)=r.
Letting limsupn→∞ in (2.5) and using the above limits with the continuity of ψ and the lower semicontinuity of ϕ, we get ψ(r)≤ψ(r)-ϕ(r), which implies that ϕ(r)=0, so r=0 by a property of ϕ. Thus, (2.4) is proved.
Step  2. We shall prove that
(2.11)lim⁡n→∞d(xn,xn+2)=0.
By (2.2), we have
(2.12)ψ(d(xn+2,xn))=ψ(d(Txn+1,Txn-1))≤ψ(M(xn+1,xn-1))-ϕ(M(xn+1,xn-1))≤ψ(M(xn+1,xn-1))
which implies that
(2.13)d(xn+2,xn)≤M(xn+1,xn-1)∀n≥1,
where
(2.14)M(xn+1,xn-1)=max⁡⁡{d(xn+1,xn-1),d(xn+1,Txn+1),d(xn-1,Txn-1)}=max⁡⁡{d(xn+1,xn-1),d(xn+1,xn+2),d(xn-1,xn)}=max⁡⁡{d(xn+1,xn-1),d(xn-1,xn)}.
Set αn=d(xn+2,xn) and βn=d(xn,xn+1). Thus, by (2.12), one can write
(2.15)ψ(αn)≤ψ(max⁡{αn-1,βn-1})-ϕ(max⁡{αn-1,βn-1})∀n≥1
which implies that
(2.16)αn≤max⁡{αn-1,βn-1}.
On the other hand, having in mind that the sequence {d(xn,xn+1)}={βn} is monotone nonincreasing, so
(2.17)βn≤βn-1≤max⁡{αn-1,βn-1}.
From (2.16) and (2.17), we have
(2.18)max⁡{αn,βn}≤max⁡{αn-1,βn-1}∀n≥1.
Therefore, the sequence {max{αn,βn}} is monotone nonincreasing, so it converges to some t≥0. Assume that t>0. Now, by (2.4), it is obvious that
(2.19)lim⁡sup⁡n→∞αn=lim⁡sup⁡n→∞max⁡{αn,βn}=lim⁡n→∞max⁡{αn,βn}=t.
Taking the limsupn→∞ in (2.15) and using (2.19) and the properties of ψ and ϕ, we obtain
(2.20)ψ(t)=ψ(lim⁡sup⁡n→∞αn)=lim⁡sup⁡n→∞ψ(αn)≤lim⁡sup⁡n→∞ψ(max⁡⁡{αn-1,βn-1})-lim⁡inf⁡n→∞ϕ(max⁡⁡{αn-1,βn-1})≤ψ(lim⁡n→∞⁡max⁡⁡{αn-1,βn-1})-ϕ(lim⁡n→∞⁡max⁡⁡{αn-1,βn-1})=ψ(t)-ϕ(t)
which implies that ϕ(t)=0, so t=0, a contradiction. Thus, from (2.19),
(2.21)lim⁡sup⁡n→∞αn=0,
and hence limn→∞αn=0, so (2.11) is proved.
Step  3. We claim that T has a periodic point.
We argue by contradiction. Assume thatT has no periodic point. Then, {xn} is a sequence of distinct points, that is, xn≠xm for all m≠n. We will show that, in this case, {xn} is g.m.s Cauchy. Suppose to the contrary. Then, there is a ɛ>0 such that for an integer k there exist integers m(k)>n(k)>k such that
(2.22)d(xn(k),xm(k))>ɛ.
For every integer k, let m(k) be the least positive integer exceeding n(k) satisfying (2.22) and such that
(2.23)d(xn(k),xm(k)-1)≤ɛ.
Now, using (2.22), (2.23), and the rectangular inequality (because {xn} is a sequence of distinct points), we find that
(2.24)ɛ<d(xm(k),xn(k))≤d(xm(k),xm(k)-2)+d(xm(k)-2,xm(k)-1)+d(xm(k)-1,xn(k))≤d(xm(k),xm(k)-2)+d(xm(k)-2,xm(k)-1)+ɛ.
Then, by (2.4) and (2.11), it follows that
(2.25)lim⁡k→+∞d(xn(k),xm(k))=ɛ.
Now, by rectangular inequality, we have
(2.26)d(xm(k),xn(k))≤d(xm(k),xm(k)-1)+d(xm(k)-1,xn(k)-1)+d(xn(k)-1,xn(k))d(xm(k)-1,xn(k)-1)≤d(xm(k)-1,xm(k))+d(xm(k),xn(k))+d(xn(k),xn(k)-1).
Letting k→∞ in the above inequalities, using (2.4) and (2.25), we obtain
(2.27)lim⁡k→∞d(xm(k)-1,xn(k)-1)=ɛ.
Therefore, by (2.4) and (2.27), we get that
(2.28)M(xm(k)-1,xn(k)-1)=max⁡{d(xm(k)-1,xn(k)-1),d(xm(k)-1,xm(k)),d(xn(k)-1,xn(k))}⟶ɛask⟶∞.
Applying (2.2) with x=xm(k)-1 and y=xn(k)-1, we have
(2.29)ψ(d(xm(k),xn(k)))=ψ(Txm(k)-1,Txn(k)-1)≤ψ(M(xm(k)-1,xn(k)-1))-ϕ(M(xm(k)-1,xn(k)-1)).
Letting k→∞ in the above inequality and using (2.25) and (2.28), we obtain
(2.30)ψ(ɛ)≤ψ(ɛ)-ϕ(ɛ)
which yields that ϕ(ɛ)=0, so ɛ=0, which is a contradiction.
Hence,{xn} is g.m.s Cauchy. Since (X,d) is a complete g.m.s, there exists u∈X such that xn→u. Applying (2.2) with x=xn and y=u, we obtain
(2.31)ψ(d(xn+1,Tu))=ψ(d(Txn,Tu))≤ψ(M(xn,u))-ϕ(M(xn,u))≤ψ(M(xn,u))
which implies that
(2.32)d(xn+1,Tu)≤M(xn,u),
where
(2.33)M(xn,u)=max⁡{d(xn,u),d(xn,xn+1),d(u,Tu)}.
Since limn→∞d(xn,u)=limn→∞d(xn,xn+1)=0, so we obtain that
(2.34)lim⁡n→∞M(xn,u)=d(u,Tu).
It follows that
(2.35)lim⁡sup⁡n→∞d(xn+1,Tu)≤d(u,Tu).
Next, we shall find a contradiction of the fact that T has no periodic point in each of the two following cases.(i)
If, for alln≥2, xn≠u and xn≠Tu, then by rectangular inequality
(2.36)d(u,Tu)≤d(u,xn)+d(xn,xn+1)+d(xn+1,Tu),and, using (2.4), we get that
(2.37)d(u,Tu)≤lim⁡sup⁡n→∞d(xn+1,Tu).From (2.35) and (2.37),
(2.38)lim⁡sup⁡n→∞d(xn+1,Tu)=d(u,Tu).Taking thelimsupn→∞ in (2.31) and using (2.34), (2.38), and the properties of ψ and ϕ, we obtain
(2.39)ψ(d(u,Tu))≤ψ(d(u,Tu))-ϕ(d(u,Tu))which implies thatd(u,Tu)=0, so u=Tu, that is, u is a fixed point of T, so u is a periodic point of T. It contradicts the fact that T has no periodic point.(ii)
Let for someq≥2, xq=u or xq=Tu. Since T has no periodic point, then obviously u≠x0. Indeed, if xq=u=x0,  so Tqx0=x0, that is, x0 is a periodic point of T, while if xq=Tu and x0=u,  so Tx0=Tu=xq=Tqx0=Tq-1(Tx0), that is, Tx0 is a periodic point of T.
For alln≥0, we have
(2.40)d(Tnu,u)=d(Tnxq,u)=d(xn+q,u)ord(Tnu,u)=d(Tn-1Tu,u)=d(Tn-1xq,u)=d(xn+q-1,u).
In the two precedent identities, the integer q≥2 is fixed, and so {xn+q} and {xn+q-1} are subsequences from {xn}, and since {xn} g.m.s. converges to u in (X,d) which is assumed to be Hausdorff, so the two subsequences g.m.s. converge to same unique limit u, that is,
(2.41)lim⁡n→∞d(xn+q,u)=lim⁡n→∞d(xn+q-1,u)=0.
Thus,
(2.42)lim⁡n→∞d(Tnu,u)=0.
Again, since (X,d) is Hausdorff, then by (2.42),
(2.43)lim⁡n→∞d(Tn+2u,u)=0.
On the other hand, since T has no periodic point, it follows that
(2.44)Tsu≠Truforanys,r∈N,s≠r.
Using (2.44) and the rectangular inequality, we may write
(2.45)|d(Tn+1u,Tu)-d(u,Tu)|≤d(Tn+1u,Tn+2u)+d(Tn+2u,u).
Letting n→∞ in the above limit and proceeding as (2.4) (since the point x0 is arbitrary), using (2.43), we obtain
(2.46)lim⁡n→∞d(Tn+1u,Tu)=d(u,Tu).
Now, by (2.2),
(2.47)ψ(d(Tn+1u,Tu))≤ψ(M(Tnu,u))-ϕ(M(Tnu,u)),
where
(2.48)M(Tnu,u)=max⁡{d(Tnu,u),d(Tnu,Tn+1u),d(u,Tu)}⟶d(u,Tu)asn⟶∞.
Letting n→∞ in (2.47) and using (2.46) and the above limit, we get that
(2.49)ψ(d(u,Tu))≤ψ(d(u,Tu))-ϕ(d(u,Tu))
which holds only if d(u,Tu)=0, that is, Tu=u, which implies that u is a periodic point of T. This contradicts the fact that T has no periodic point.
Consequently,T admits a periodic point, that is, there exists u∈X such that u=Tpu for some p≥1.
Step  4. Existence of a fixed point of T.
Ifp=1, then u=Tu, that is, u is a fixed point of T. Suppose now that p>1. We will prove that a=Tp-1u is a fixed point of T. Suppose that it is not the case, that is, Tp-1u≠Tpu. Then, d(Tp-1u,Tpu)>0 and ϕ(d(Tp-1u,Tpu))>0, which implies that ϕ(M(Tp-1u,Tpu))>0. Now, using inequality (2.2), we obtain
(2.50)ψ(d(u,Tu))=ψ(d(Tpu,Tp+1u))=ψ(d(T(Tp-1u),T(Tpu)))≤ψ(M(Tp-1u,Tpu))-ϕ(M(Tp-1u,Tpu))<ψ(M(Tp-1u,Tpu))
which by the monotone nondecreasing property of ψ implies
(2.51)d(u,Tu)<M(Tp-1u,Tpu),
where
(2.52)M(Tp-1u,Tpu)=max⁡⁡{d(Tp-1u,Tpu),d(Tp-1u,Tpu),d(Tpu,Tp+1u)}=max⁡⁡{d(Tp-1u,Tpu),d(u,Tu)}=d(Tp-1u,Tpu)
because otherwise we get a contradiction with (2.51). Thus, (2.51) becomes
(2.53)d(u,Tu)<d(Tp-1u,Tpu).
Again, using (2.2), we have
(2.54)ψ(d(Tp-1u,Tpu))=ψ(d(T(Tp-2u),T(Tp-1u)))≤ψ(M(Tp-2u,Tp-1u))-ϕ(M(Tp-2u,Tp-1u))<ψ(M(Tp-2u,Tp-1u)).
Again, this implies that
(2.55)d(Tp-1u,Tpu)<M(Tp-2u,Tp-1u),
Where
(2.56)M(Tp-2u,Tp-1u)=max⁡⁡{d(Tp-2u,Tp-1u),d(Tp-2u,Tp-1u),d(Tp-1u,Tpu)}=max⁡⁡{d(Tp-2u,Tp-1u),d(Tp-1u,Tpu)}=d(Tp-2u,Tp-1u)
because of (2.55). Thus, from (2.55),
(2.57)d(Tp-1u,Tpu)<d(Tp-2u,Tp-1u).
Continuing this process as (2.53) and (2.57), we find that
(2.58)d(u,Tu)<d(Tp-1u,Tpu)<d(Tp-2u,Tp-1u)<⋯<d(u,Tu)
which is a contradiction. We deduce that a=Tp-1u is a fixed point of T.
Step  5. Uniqueness of the fixed point of T.
Suppose that there are two distinct pointsb,c∈X such that Tb=b and Tc=c. Then, M(b,c)=max{d(b,c),d(b,Tb),d(c,Tc)}=d(b,c) and ϕ(d(b,c))>0. By (2.2), we obtain
(2.59)ψ(d(b,c))=ψ(d(Tb,Tc))≤ψ(M(b,c))-ϕ(M(b,c))=ψ(d(b,c))-ϕ(d(b,c))<ψ(d(b,c))
a contradiction. Thus, T has a unique fixed point. This completes the proof of Theorem 2.2.Now, we state some corollaries of Theorem2.2, which are given in the following.Corollary 2.3.
Let(X,d) be a Hausdorff and complete generalized metric space. Suppose that T:X→X is such that, for all x,y∈X, there exists k∈[0,1) and
(2.60)d(Tx,Ty)≤kmax⁡{d(x,y),d(x,Tx),d(y,Ty)},
then T has a unique fixed point.Proof.
It suffices to takeψ(t)=t and ϕ(t)=(1-k)t in Theorem 2.2.Corollary 2.4.
Let(X,d) be a Hausdorff and complete generalized metric space. Suppose that T:X→X is such that, for all x,y∈X, there exists α∈[0,1/2) and
(2.61)(d(Tx,Ty))≤α[d(x,Tx)+d(y,Ty)],
then T has a unique fixed point.Proof.
Letk=2α, so k∈[0,1). Also, if (2.61) holds, so
(2.62)(d(Tx,Ty))≤α[d(x,Tx)+d(y,Ty)]=kd(x,Tx)+d(y,Ty)2≤kmax⁡⁡{d(x,y),d(x,Tx),d(y,Ty)}.
Then, it suffices to apply Corollary 2.3.Another easy consequence of Corollary2.3 (a Reich contraction type) is the following.Corollary 2.5.
Let(X,d) be a Hausdorff and complete generalized metric space. Suppose that T:X→X is such that, for all x,y∈X, there exists k∈[0,1/3) and
(2.63)(d(Tx,Ty))≤k[d(x,y)+d(x,Tx)+d(y,Ty)],
then T has a unique fixed point.Corollary 2.6.
LetT satisfy the conditions of Theorem 2.2, except that condition (2.2) is replaced by the following: there exist positive Lebesgue integrable functions u and v on ℝ+ such that ∫0ɛ‍u(t)dt>0 and ∫0ɛ‍v(t)dt>0 for each ɛ>0 and that
(2.64)∫0ψ(d(Tx,Ty))u(t)dt≤∫0ψ(M(x,y))u(t)dt-∫0ϕ(M(x,y))v(t)dt.
Then, T has a unique fixed point.Proof.
Consider the functions(2.65)φ0(x)=∫0xu(t)dt,φ1(x)=∫0xv(t)dt.
Then, (2.64) becomes
(2.66)(φ0∘ψ)(d(Tx,Ty))≤(φ0∘ψ)(M(x,y))-(φ1∘ϕ)(M(x,y)),
And, putting ψ0=φ0∘ψ and ϕ0=φ1∘ϕ and applying Theorem 2.2, we obtain the proof of Corollary 2.6 (it is easy to verify that ψ0∈Ψ and ϕ0∈Φ).Corollary 2.7.
Let(X,d) be a Hausdorff and complete generalized metric space. Let T:X→X. Assume there exist positive Lebesgue integrable functions u and v on ℝ+ such that ∫0ɛ‍u(t)dt>0 and ∫0ɛ‍v(t)dt>0 for each ɛ>0 and for all x,y∈X, and
(2.67)∫0d(Tx,Ty)u(t)dt≤∫0M(x,y)u(t)dt-∫0M(x,y)v(t)dt,
then T has a unique fixed point.Proof.
It follows by takingψ(t)=ϕ(t)=t in Corollary 2.6.Corollary 2.8.
Let(X,d) be a Hausdorff and complete generalized metric space. Let T:X→X. Assume there exist k∈[0,1) and a positive Lebesgue integrable function u on ℝ+ such that ∫0ɛ‍u(t)dt>0 for each ɛ>0 and for all x,y∈X, and
(2.68)∫0d(Tx,Ty)u(t)dt≤k∫0max⁡⁡{d(x,y),d(x,Tx),d(y,Ty)}u(t)dt,
then T has a unique fixed point.Proof.
It suffices to takev(t)=(1-k)u(t) in Corollary 2.7.Finally, let us finish this paper by noticing the following remark.Remark 2.9.
(i) Theorem2.2 extends Theorem 3.1 of Lakzian and Samet [9].
(ii) Corollary2.3 extends the results of Branciari [1], Azam and Arshad [2], and Sarma et al. [13].
(iii) Corollary2.8 extends Theorem 2 of Samet [11].
(iv) Several publications attempting to generalize fixed-point theorems in metric spaces to g.m.s are plagued by the use of some false properties given in [1] (see, e.g., [2–5]). This was observed by Das and Dey [7] who proved a fixed-point theorem without using the false properties. Subsequently, but independently, this was also observed by Samet [12] and Sarma et al. [13] who proved fixed-point theorems assuming that the generalized metric space is Hausdorff. Here, we give a rigorous proof of Theorem 2.2 by taking the same assumption.



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