In this paper, the notion of L-contractions is introduced and a new fixed point theorem for such contractions is established.

Hanseo University1. Introduction and Preliminaries

Branciari [1] introduced the notion of generalized metric spaces and obtained a generalization of the Banach contraction principle, whereafter many authors proved various fixed point results in such spaces, for example, [2–8] and references therein. Also, Suzuki et al. [9] and Abtahi et al. [10] studied ν-generalized metric spaces and proved the Banach and Kannan contraction principles in such spaces, and Mitrović et al. [11] introduced the notion of bν(s)-generalized metric spaces and Banach and Reich contraction principles in such spaces.

In particular, Jleli and Samet [12] introduced the notion of θ-contractions and gave a generalization of the Banach contraction principle in generalized metric spaces, where θ:(0,∞)→(1,∞) is a function satisfying the following conditions:

θ is nondecreasing;

∀{tn}⊂(0,∞), (1)limn→∞θtn=1⇔limn→∞tn=0+;

∃r∈(0,1)∧l∈(0,∞): (2)limt→0+θt-1tr=l.

Also, Ahmad et al. [13] extended the result of Jleli and Samet [12] to metric spaces by applying the following simple condition (θ4) instead of (θ3).

θ is continuous on (0,∞).

Recently, Khojasteh et al. [14] introduced the notion of Z-contractions by defining the concept of simulation functions. They unified some existing metric fixed point results. Afterward, many authors ([15–19] and references therein) obtained generalizations of the result of [14].

In the paper, we introduce the concept of a new type of contraction maps, and we establish a new fixed point theorem for such contraction maps in the setting of generalized metric spaces.

Let L be the family of all mappings ξ:[1,∞)×[1,∞)→R such that

ξ(1,1)=1;

ξ(t,s)<s/t∀s,t>1;

for any sequence {tn},{sn}⊂(1,∞) with tn≤sn∀n=1,2,3,⋯(3)limn→∞tn=limn→∞sn>1⇒limn→∞supξtn,sn<1.

We say that ξ∈L is a L-simulation function.

Note that ξ(t,t)<1∀t>1.

Example 1.

Let ξb,ξw,ξ:[1,∞)×[1,∞)→R be functions defined as follows, respectively:

ξb(t,s)=sk/t∀t,s≥1, where k∈(0,1);

ξw(t,s)=s/tϕ(s)∀t,s≥1, where ϕ:[1,∞)→[1,∞) is nondecreasing and lower semicontinuous such that ϕ-1({1})=1;(4)ξt,s=1ifs,t=1,1,s2tifs<t,sλtotherwise,

∀s,t≥1, where λ∈(0,1).

Then ξb,ξw,ξ∈L.

We recall the following definitions which are in [1].

Let X be a nonempty set, and let d:X×X→[0,∞) be a map such that for all x,y∈X and for all distinct points u,v∈X, each of them is different from x and y:

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

d(x,y)=d(y,x);

d(x,y)≤d(x,u)+d(u,v)+d(v,y).

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

Let (X,d) be a generalized metric space, let {xn}⊂X be a sequence, and x∈X.

Then we say that

{xn} is convergent to x (denoted by limn→∞xn=x) if and only if limn→∞d(xn,x)=0;

{xn} is Cauchy if and only if limn,m→∞d(xn,xm)=0;

(X,d) is complete if and only if every Cauchy sequence in X is convergent to some point in X.

Let (X,d) be a generalized metric space.

A map T:X→X is called continuous at x∈X if, for any V∈τ containing Tx, there exists U∈τ containing x such that TU⊂V, where τ is the topology on X induced by the generalized metric d. That is,(5)τ=U⊂X:∀x∈U,∃B∈β,x∈B⊂U,β=Bx,r:x∈X,∀r>0,Bx,r=y∈X:dx,y<r.

If T is continuous at each point x∈X, then it is called continuous.

Note that T is continuous if and only if it is sequentially continuous, i.e., limn→∞d(Txn,Tx)=0 for any sequence {xn}⊂X with limn→∞d(xn,x)=0.

Remark 2 (see [<xref ref-type="bibr" rid="B23">6</xref>]).

If d is a generalized metric on X, then it is not continuous in each coordinate.

Lemma 3 (see [<xref ref-type="bibr" rid="B13">20</xref>]).

Let (X,d) be a generalized metric space, let {xn}⊂X be a Cauchy sequence, and x,y∈X. If there exists a positive integer N such that

xn≠xm∀n,m>N;

xn≠x∀n>N;

xn≠y∀n>N;

limn→∞d(xn,x)=limn→∞d(xn,y),

then x=y.

2. Fixed Point Theorems

We denote by Θ the class of all functions θ:(0,∞)→(1,∞) such that conditions (θ1) and (θ2) hold.

A mapping T:X→X is called L-contraction with respect to ξ if there exist θ∈Θ and ξ∈L such that, for all x,y∈X with d(Tx,Ty)>0, (6)ξθdTx,Ty,θdx,y≥1.

Note that if T is L-contraction with respect to ξ, then it is continuous. In fact, let x∈X be a point and let {xn}⊂X be any sequence such that (7)limn→∞dxn,x=0+,dTxn,Tx>0∀n=1,2,3,⋯.

Then from (θ2) limn→∞θ(d(xn,x))=1.

It follows from (6) and (ξ2) that (8)1≤ξθdTxn,Tx,θdxn,x<θdxn,xθdTxn,Tx, which implies (9)θdTxn,Tx<θdxn,x.

Since θ is nondecreasing, we have (10)dTxn,Tx<dxn,x,

and so (11)limn→∞dTxn,Tx=0.Hence T is continuous.

Now, we prove our main result.

Theorem 4.

Let (X,d) be a complete generalized metric space, and let T:X→X be a L-contraction with respect to ξ.

Then T has a unique fixed point, and for every initial point x0∈X, the Picard sequence {Tnx0} converges to the fixed point.

Proof.

Firstly, we show uniqueness of fixed point whenever it exists.

Assume that w and u are fixed points of T.

If u≠z, then d(w,u)>0, and so it follows from (6) that(12)1≤ξθdTw,Tu,θdw,u=ξθdw,u,θdw,u<θdw,uθdw,u. Hence (13)θdw,u<θdw,u which is a contradiction.

Hence w=u, and fixed point of T is unique.

Secondly, we prove existence of fixed point.

Let x0∈X be a point. Define a sequence {xn}⊂X by xn=Txn-1=Tnx0∀n=1,2,3⋯.

If xn0=xn0+1 for some n0∈N, then xn0 is a fixed point of T, and the proof is finished.

Assume that (14)xn-1≠xn∀n=1,2,3⋯.

It follows from (6) and (14) that ∀n=1,2,3,⋯(15)1≤ξθdTxn-1,Txn,θdxn-1,xn=ξθdxn,xn+1,θdxn-1,xn<θdxn-1,xnθdxn,xn+1.

Consequently, we obtain that (16)θdxn,xn+1<θdxn-1,xn∀n=1,2,3,⋯ which implies (17)dxn,xn+1<dxn-1,xn∀n=1,2,3,⋯.

Hence {d(xn-1,xn)} is a decreasing sequence, and so there exists r≥0 such that (18)limn→∞dxn-1,xn=r.

We now show that r=0.

Assume that r≠0.

Then it follows from (θ2)that(19)limn→∞θdxn-1,xn≠1,and so(20)limn→∞θdxn-1,xn>1.

Let sn=θ(d(xn-1,xn)) and tn=θ(d(xn,xn+1))∀n=1,2,3,⋯.

From (ξ3) we obtain (21)1≤limn→∞supξtn,sn<1 which is a contradiction.

Thus we have (22)limn→∞dxn-1,xn=0and so (23)limn→∞θdxn-1,xn=1.

We show that (24)limn→∞dxn-1,xn+1=0.

We consider three cases.

Case 1. xn≠xn+2∀n=1,2,3,⋯.

From (6) and (14) we obtain that ∀n=1,2,3,⋯(25)1≤ξθdTxn-1,Txn+1,θdxn-1,xn+1=ξθdxn,xn+2,θdxn-1,xn+1<θdxn-1,xn+1θdxn,xn+2, and so (26)θdxn,xn+2<θdxn-1,xn+1∀n=1,2,3,⋯ which implies (27)dxn,xn+2<dxn-1,xn+1∀n=1,2,3,⋯. Hence {d(xn-1,xn+1)} is decreasing.

In a manner similar to that which proved (22), we have (28)limn→∞dxn-1,xn+1=0.

Case 2. There exists n0≥1 such that xn0=xn0+2.

From the first term to the n0 th term shall be removed, and let xn=xn0+n∀n=1,2,3,⋯.

Then xn≠xn+2∀n=1,2,3,⋯. By Case 1, we have (29)limn→∞dxn-1,xn+1=0.

Case 3. xn=xn+2∀n=0,1,2,⋯.

We have (30)dxn-1,xn+1=0∀n=1,2,3,⋯.

Hence (31)limn→∞dxn-1,xn+1=0.

In all cases, (24) is satisfied.

Now, we show that {xn} is bounded.

If {xn} is not bounded, then there exists a subsequence {xn(k)} of {xn} such that n1=1 and ∀k=1,2,3,…,n(k+1) is the minimum integer greater than n(k) with (32)dxnk+1,xnk>1,dxl,xnk≤1for n(k)≤l≤n(k+1)-1.

Then we have(33)1<dxnk+1,xnk≤dxnk+1,xnk+1-2+dxnk+1-2,xnk+1-1+dxnk+1-1,xnk≤dxnk+1,xnk+1-2+dxnk+1-2,xnk+1-1+1.

By letting k→∞ in the above, we obtain (34)limk→∞dxnk+1,xnk=1.

By using (22), (34), and condition (d3), we deduce that (35)limk→∞dxnk+1-1,xnk-1=1.

It follows from (θ2), (34), and (35) that (36)limk→∞θdxnk+1,xnk>1,(37)limn→∞θdxnk+1-1,xnk-1>1.

From (6) and (32) we infer that(38)1≤ξθdTxnk+1-1,Txnk-1,θdxnk+1-1,xnk-1=ξθdxnk+1,xnk,θdxnk+1-1,xnk-1<θdxnk+1-1,xnk-1θdxnk+1,xnkwhich implies (39)θdxnk+1,xnk<θdxnk+1-1,xnk-1.

Let (40)sn=θ(dxnk+1-1,xnk-1,tn=θdxnk+1,xnk∀n=1,2,3,⋯.

Then tk<sk∀n=1,2,3,⋯ and limk→∞tk=limk→∞sk>1.

It follows from (ξ3) that (41)1≤limk→∞supξtk,sk<1 which is a contradiction.

Thus {xn} is bounded.

Now, we show that {xn} is a Cauchy sequence.

Let (42)Mn=supdxi,xj:i,j≥n.

Clearly, (43)0≤Mn+1≤Mn<∞∀n=1,2,3,⋯and so there exists M≥0 such that (44)limn→∞Mn=M.

Assume that M>0.

It follows from (42) that ∀k=1,2,3,⋯ there exist n(k),m(k)≥k with (45)Mk-1k<dxmk,xnk≤Mk.

So (46)limk→∞dxmk,xnk=limk→∞Mk=M.

It follows from (6) and (14) that(47)ξθdTxmk-1,Txnk-1,θdxmk-1,xnk-1=ξθdxmk,xnk,θdxmk-1,xnk-1<θdxmk-1,xnk-1θdxmk,xnk which implies (48)θdxmk,xnk<θdxmk-1,xnk-1. Hence we have(49)dxmk,xnk<dxmk-1,xnk-1≤dxmk-1,xmk+dxmk,xnk+dxnk,xnk-1.

Letting k→∞ in the above inequality, we obtain (50)limn→∞dxmk-1xnk-1=M.

Let (51)sk=θdxmk-1,xnk-1,tk=θdxmk,xnk∀k=1,2,3,⋯.

Then tk<sk∀k=1,2,3,⋯. Since M>0, (52)limk→∞tk=limk→∞sk>1.

Thus we have (53)1≤limk→∞supξtk,sk<1 which is a contradiction.

Hence M=0, and hence {xn} is a Cauchy sequence.

Since X is complete, there exists z∈X such that (54)limn→∞dxn,z=0.

Because T is continuous, (55)limn→∞dxn,Tz=limn→∞dTxn-1,Tz=0.

By Lemma 3, z=Tz.

We give an example to illustrate Theorem 4.

Example 5.

Let X={1,2,3,4} and define d:X×X→[0,∞) as follows:(56)d1,2=d2,1=3,d2,3=d3,2=d1,3=d3,1=1,d1,4=d4,1=d2,4=d4,2=d3,4=d4,3=4,dx,x=0∀x∈X.

Then (X,d) is a complete generalized metric space, but not a metric space (see [21]).

Define a map T:X→X by (57)Tx=3x≠4,1x=4. And define a function θ:(0,∞)→(1,∞) by (58)θt=et.

We now show that T is a L-contraction with respect to ξb, where ξb(t,s)=sk/t∀t,s≥1, k=1/2.

We have (59)dTx,Ty=d1,3=1x=4,y≠4,d1,1=0x=4,y=4,d3,3=0x≠4,y≠4 so (60)dTx,Ty>0⇔x=4,y≠4.

We have, for x=4 and y≠4, (61)dx,y=4,dTx,Ty=1.

We deduce that, for all x,y∈X with d(Tx,Ty)>0,(62)ξbθdTx,Ty,θdx,y=θdx,ykθdTx,Ty=e41/2e1=e>1.

Thus all hypotheses of Theorem 4 are satisfied, and T has a fixed point x∗=3.

Note that Banach’s contraction principle is not satisfied with the usual metric ρx,y=x-y∀x,y∈X. In fact, if x=2, y=4, then (63)ρT2,T4≤kρ2,4,k∈0,1 which implies (64)k≥1.

Also, note that the θ-contraction condition [13] does not hold.

Let θ(t)=et,∀t>0.

Then θ(t) satisfies conditions (θ1), (θ2), and (θ4).

If (65)θρT2,T4≤θρ2,4k,wherek∈0,1 then (66)e2≤e2k and so k≥1. Hence T is not θ-contraction map.

By taking ξ=ξb in Theorem 4, we obtain Corollary 6.

Corollary 6.

Let (X,d) be a complete generalized metric space, and let T:X→X be a mapping such that for all x,y∈X with d(Tx,Ty)>0(67)θdTx,Ty≤θdx,yk where θ∈Θ and k∈(0,1).

Then T has a unique fixed point.

Remark 7.

Corollary 6 is a generalization of Theorem 2.1 of [12] without condition (θ3) and Theorem 2.2 of [13] without condition (θ4).

By taking ξ=ξw in Theorem 4, we obtain Corollary 8.

Corollary 8.

Let (X,d) be a complete generalized metric space, and let T:X→X be a mapping such that for all x,y∈X with d(Tx,Ty)>0(68)θdTx,Ty≤θdx,yϕθdx,y where θ∈Θ and ϕ:[1,∞)→[1,∞) is nondecreasing and lower semicontinuous such that ϕ-1({1})=1.

Then T has a unique fixed point.

Corollary 9.

Let (X,d) be a complete generalized metric space, and let T:X→X be a mapping such that for all x,y∈X with d(Tx,Ty)>0(69)dTx,Ty≤dx,y-φdx,ywhere φ:[0,∞)→[0,∞) is nondecreasing and lower semicontinuous such that φ-1({0})=0.

Then T has a unique fixed point.

Proof.

Condition (69) implies T is continuous.

Let θ(t)=et,∀t>0.

From (69) we have that, for all x,y∈X with d(Tx,Ty)>0,(70)θdTx,Ty=edTx,Ty≤edx,y-φdx,y=edx,yeφdx,y.

Let φ(t)=ln(ϕ(θ(t))),∀t≥0, where ϕ:[1,∞)→[1,∞) is nondecreasing and lower semicontinuous such that ϕ-1({1})=1.

Then φ is nondecreasing and lower semicontinuous, and (71)φt=0⇔ϕθt=1⇔θt=et=1⇔t=0.

It follows from (70) that, for all x,y∈X with d(Tx,Ty)>0, (72)θdTx,Ty≤θdx,yelnϕθdx,y=θdx,yϕθdx,y.

By Corollary 8, T has a unique fixed point.

By taking θ(t)=2-2/πarctan(1/tα), where α∈(0,1),t>0 in Corollary 8, we obtain the following result.

Corollary 10.

Let (X,d) be a complete generalized metric space, and let T:X→X be a mapping such that for all x,y∈X with d(Tx,Ty)>0(73)2-2πarctan1dTx,Tyα≤2-2/πarctan1/dx,yαϕ2-2/πarctan1/dx,yα where α∈(0,1) and ϕ:[1,∞)→[1,∞) is nondecreasing and lower semicontinuous such that ϕ-1({1})=1.

Then T has a unique fixed point.

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The author express his gratitude to the referees for careful reading and giving variable comments. This research was supported by Hanseo University.

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