# New Contractive Mappings and Their Fixed Points in Branciari Metric Spaces

Authors: Hayel N. Saleh; Mohammad Imdad; Thabet Abdeljawad; Mohammad Arif
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we introduce the notion of generalizedL-contractions which enlarge the class of ℒ-contractions initiated by Cho in 2018. Thereafter, we also, define the notion of L∗-contractions. Utilizing our newly introduced notions, we establish some new fixed-point theorems in the setting of complete Branciari’s metric spaces, without using the Hausdorff assumption. Moreover, some examples and applications to boundary value problems of the fourth-order differential equations are given to exhibit the utility of the obtained results.


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## Body

## 1. Introduction

In 2000, Branciari [1] initiated the concept of a generalized metric by replacing the natural triangle inequality of a metric with a relatively more general inequality termed as rectangular (or quadrilateral) inequality which involves four points instead of three. In the literature, such a metric is known as the Branciari metric. Branciari [1] assumed that each of the Branciari metric space becomes a Hausdorff topological space, and the Branciari metric is a continuous function in each coordinate. Sarma et al. [2] and Samet [3] showed that these assumptions were not correct. Meanwhile, several authors obtained various fixed-point results in the Branciari metric space with the assumption that the space is a Hausdorff and (or) the Branciari metric is continuous. However, it was shown (e.g. [4, 5]) that in general, neither the Hausdorff topological property nor the continuity of the metric is required in the proofs. For a recent update or extension of the Branciari metric spaces, we refer to [6].On the other hand, in 2014, Jleli and Samet [7] introduced the concept of Θ-contractions and proved some fixed-point results in the setting of the Branciari metric spaces. Very recently, Cho [8] introduced the notion of L-contractions which unify several concepts of contractions in the existing literature including Θ-contractions in the setting of the Branciari metric spaces. For more notions and results in such spaces, we refer the reader to [9–17].This paper is aimed at introducing two types of contraction mappings, namely, generalizedL-contractions and L∗-contractions. For both types of contractions, we prove separate fixed-point theorems in the setting of the complete Branciari metric spaces. The obtained results extend, generalize, and improve some results of the existing literature. Moreover, applications to boundary value problems of the fourth-order are given to exhibit the utility of the obtained results.


## 2. Preliminaries

The following definitions and basic results are needed in the sequel.Definition 1 [1].
LetX be a nonempty set and d:X×X⟶0,∞ a mapping such that for all x,y∈X and all z≠w∈X/x,y:
(BMS1)
dx,y=0, if and only if x=y(BMS2)
dx,y=dy,x(BMS3)
dx,y≤dx,z+dz,w+dw,y.The metricd is called a Branciari metric, and the pair (X,d) is called a Branciari metric space.The next example shows that the topologies of the Branciari metric spaces and the usual metric spaces are different. In particular, we have the following:(i)
The Branciari metric need not be continuous in both variables(ii)
The Branciari metric space is not necessarily Hausdorff(iii)
An open ball need not be an open set(iv)
A convergent sequence is not necessarily a Cauchy sequence.Example 2 [2].
LetX=A∪B, where A=0,2 and B=1/n:n∈ℕ. Define d:X×X→0,∞ as
(1)dx,y=0,ifx=y,1,ifx≠yandx,y⊂Aorx,y⊂B,y,ifx∈Aandy∈B,x,ifx∈Bandy∈A.
ThenX,d is a complete Branciari metric space. However, it is easy to see that
(i)
limn→∞d1/n,1/2≠d0,1/2 although limn→∞1/n=0, and hence the function d is not continuous(ii)
There does not existr>0 such that Br0∩Br2=ϕ, and hence the respective topology is not a Hausdorff(iii)
B2/31/3=0,2,1/3; however, there does not exist r>0 such that Br0⊆B2/31/3, and hence an open ball need not be an open set(iv)
The sequence1/nn∈ℕ converges to both 0 and 2, and it is not a Cauchy sequence.Definition 3 [7].
LetX,d be a Branciari metric space. A mapping T:X⟶X is said to be a Θ-contraction if there exist Θ∈Ω1,2,3 and k∈0,1 such that (for all x,y∈X)
(2)dTx,Ty>0⇒ΘdTx,Ty≤Θdx,yk,where Ω1,2,3 is the family of all functions Θ: 0,∞→1,∞ which satisfy the following conditions:
(Θ1) Θ is nondecreasing
(Θ2) For each sequence αn⊂0,∞, limn→∞Θαn=1⇔limn→∞αn=0+
(Θ3) There exists r∈0,1 and l∈0,∞ such that limα→0+Θα−1/αr=l.Theorem 4 [7].
LetX,d be a complete Branciari metric space and T:X⟶X a Θ-contraction mapping. Then T has a unique fixed point.Imdad et al. [18] observed that this theorem can be proved without the condition (Θ1). Also, Ahmad et al. [19] replaced the condition (Θ3) by the following one:Θ4Θ is continuous.Remark 5.
It is known that everyΘ-contraction mapping is continuous.In the sequel, we adopt the following notations:(i)
Ω1,2,3 is the class of all functions Θ which satisfy Θ1-Θ3(ii)
Ω1,2,4 is the class of all functions Θ which satisfy Θ1, Θ2, and Θ4(iii)
Ω1,2,3,4 is the class of all functions Θ which satisfy Θ1-Θ4.In 2018, Cho [8] introduced the notion of L-contractions which unify several concepts of contractions in the existing literature including Θ-contractions as under one unifying concept:Definition 6 [8].
LetX,d be a Branciari metric space. A mapping T:X⟶X is said to be an L-contraction with respect to ζ∈L if there exists Θ∈Ω1,2,4 such that (for all x,y∈X)
(3)dTx,Ty>0⇒ζΘdTx,Ty,Θdx,y≥1,where L is the class of all functions ζ:1,∞×1,∞⟶ℝ which satisfy the following conditions ζ1∗:
(ζ1∗) ζ1,1=1
(ζ2∗) ζt,s<s/t, for all t,s>1
(ζ3∗) If tn and sn are two sequences in 1,∞ with tn<sn, such that limn→∞tn=limn→∞sn>1, then limsupn→∞ζtn,sn<1.Example 7 [8].
Letζk,ζψ:1,∞×1,∞⟶ℝ be two functions defined as under:
(a)
ζkt,s=sk/t, for all t,s≥1, where k∈0,1(b)
ζψt,s=s/tψs, for all t,s≥1, where ψ:1,∞⟶1,∞ is a lower semicontinuous and nondecreasing function with ψ−11=1
Thenζk,ζψ∈L.Example 8.
Letζψ,φ:1,∞×1,∞⟶ℝ be a function defined by
(4)ζψ,φt,s=φsψs,∀t,s≥1,where ψ,φ:1,∞⟶1,∞ are upper semicontinuous from the right such that φt<t≤ψt, for all t>1. Then ζψ,φ∈L.Based on the above definition, the author in [8] proves the following theorem.Theorem 9 [8].
LetX,d be a complete Branciari metric space and T:X⟶X an L-contraction mapping. Then T has a unique fixed point.Remark 10.
It is known that everyL-contraction mapping is continuous.Remark 11.
Letan,bn,cn be three sequences of real numbers such that limn→∞an=a, limn→∞bn=b, and limn→∞cn=c. Then
(i)limn→∞maxan,bn,cn=maxa,b,c
(ii)limn→∞minan,bn,cn=mina,b,c.The following lemmas are useful in the sequel.Lemma 12 [20].
Letxn be a Cauchy sequence in a Branciari metric space X,d such that xn≠xm whenever n≠m. Then xn can converge to at most one point.Lemma 13 [21].
Letxn be a Cauchy sequence in a Branciari metric space X,d such that limn→∞dxn,x=0, for some x∈X. Then, limn→∞dxn,y=dx,y, for all y∈X. In particular, xn does not converge to y if y≠x.


## 3. Fixed-Point Results for GeneralizedL-Contractions

We begin this section by introducing the concept of generalizedL-contractions followed by the main result of this section as follows:Definition 14.
LetX,d be a Branciari metric space and T:X⟶X. Then T is said to be a generalized L-contraction with respect to ζ∈L if there exist Θ∈Ω1,2,4 and constant L≥0 such that (for all x,y∈X)
(5)dTx,Ty>0⇒ζΘdTx,Ty,Θdx,y+LNx,y≥1,where Nx,y=mindx,Ty,dy,Tx,dx,Tx+dy,Ty/2.Theorem 15.
LetX,d be a complete Branciari metric space and T:X⟶X. If T is a continuous generalized L-contraction mapping with respect to ζ∈L, then it has a unique fixed point.Proof.
Letx0 be an arbitrary element of X and define a Picard sequence xn⊆X by xn=Txn−1, for all n∈ℕ. If dxn0,xn0+1=0, for some n0∈ℕ, then xn0 is a fixed point of T and the proof is finished. Now, assume that dxn,xn+1>0, for all n∈ℕ. Using the contractive condition (5) and ζ2∗, we have
(6)1≤ζΘdTxn−1,Txn,Θdxn−1,xn+LNxn−1,xn=ζΘdxn,xn+1,Θdxn−1,xn+LNxn−1,xn<Θdxn−1,xn+LNxn−1,xnΘdxn,xn+1,which implies that
(7)Θdxn,xn+1<Θdxn−1,xn+LNxn−1,xn,∀n∈ℕ.Notice that(8)Nxn−1,xn=mindxn−1,Txn,dxn,Txn−1,dxn−1,Txn−1+dxn,Txn2=mindxn−1,xn+1,dxn,xn,dxn−1,xn+dxn,xn+12=0.Hence, inequality (7) becomes
(9)Θdxn,xn+1<Θdxn−1,xn,∀n∈ℕ,which implies (in view of Θ1) that
(10)dxn,xn+1<dxn−1,xn,∀n∈ℕ.Therefore, the sequencedxn−1,xn is decreasing and bounded below by 0. This insures the existence of a number r≥0 such that limn→∞dxn−1,xn=r. Assume that r≠0, then it follows from Θ2 that
(11)limn→∞Θdxn−1,xn>1.Settingtn=Θdxn,xn+1 and sn=Θdxn−1,xn. In view of (9), (11), and Θ4, we have limn→∞tn=limn→∞sn>1 and tn<sn, for all n∈ℕ. Therefore, applying the condition ζ3∗, we deduce
(12)1≤limsupn→∞ζtn,sn<1,which is a contradiction, and hence we must have
(13)limn→∞dxn−1,xn=0.Now, assume thatxm=xn, for some m>n. Then, also xm+1=xn+1. Using (9), we get
(14)Θdxm,xm+1<Θdxm−1,xm<Θdxm−2,xm−1<⋯<Θdxn,xn+1=Θdxm,xm+1,which is a contradiction. Therefore, we conclude that xm≠xn, for all n≠m.Next, we claim that the sequencexn is a Cauchy sequence in X,d. On the contrary, assume that it is not Cauchy, then there exists an ε>0 for which we can find two subsequences xnk and xmk of xn such that nk>mk≥k, for all k∈ℕ and
(15)dxmk,xnk≥ε.Suppose thatnk is the least integer exceeding mk satisfying inequality (15). Then, we have
(16)dxmk,xnk−1<ε.Using (15), (16), and the rectangular inequality, we get
(17)ε≤dxmk,xnk≤dxmk,xnk−2+dxnk−2,xnk−1+dxnk−1,xnk<ε+dxnk−2,xnk−1+dxnk−1,xnk.On taking the limit ask⟶∞ and making use of (13), we obtain
(18)limk→∞dxmk,xnk=ε.Employing the rectangular inequality once again, we get(19)dxmk,xnk≤dxmk,xmk−1+dxmk−1,xnk−1+dxnk−1,xnk≤2dxmk,xmk−1+dxmk,xnk+2dxnk−1,xnk.On lettingk⟶∞ and using (13) as well as (18), we get
(20)limk→∞dxmk−1,xnk−1=ε.Now, using (5) and ζ2∗, we obtain
(21)1≤ζΘdTxmk−1,Txnk−1,Θdxmk−1,xnk−1+LNxmk−1,xnk−1=ζΘdxmk,xnk,Θdxmk−1,xnk−1+LNxmk−1,xnk−1<Θdxmk−1,xnk−1+LNxmk−1,xnk−1Θdxmk,xnk,where
(22)Nxmk−1,xnk−1=mindxmk−1,xnk,dxnk−1,xmk,dxmk−1,xmk+dxnk−1,xnk2.Consequently, we deduce that(23)Θdxmk,xnk<Θdxmk−1,xnk−1+LNxmk−1,xnk−1,∀k∈ℕ.Lettk=Θdxmk,xnk and sk=Θdxmk−1,xnk−1+LNxmk−1,xnk−1. Then in view of Remark 11 and (23), we have limk→∞tk=limk→∞sk>1 and tk<sk,∀k∈ℕ. So, on using ζ3∗, we obtain
(24)1≤limsupk→∞ζtk,sk<1,which is a contradiction. Therefore, xn must be a Cauchy sequence in X,d. Since X,d is complete, then there exists u∈X such that limt→∞xn=u, that is,
(25)limn→∞dxn,u=0.AsT is continuous, then we get that (due to (25))
(26)limn→∞dxn+1,Tu=limn→∞dTxn,Tu=0,that is, limn→∞xn+1=Tu. Using Lemma 13, we conclude that u=Tu, that is, u is a fixed point of T.Finally, we show that the fixed point of the mappingT is unique. On the contrary, assume that there are two fixed points such that du,z=dTu,Tz>0. From (5), we have
(27)1≤ζΘdTu,Tz,Θdu,z+LNu,z=ζΘdu,z,Θdu,z+LNu,z<Θdu,z+LNu,zΘdu,z,where Nu,z=mindu,Tz,dz,Tu,du,Tu+dz,Tz/2=0.This implies that(28)Θdu,z<Θdu,z,which is a contradiction. Then T has a unique fixed point.Next, we furnish the following illustrative example which shows that Theorem15 is a genuine extension of Theorems 4 and 9.Example 16.
LetX=1,2,3,4,5 and define a mapping d:X×X⟶0,∞ as
(i)
d1,2=d1,4=4(ii)
d1,5=d2,5=d3,5=1(iii)
d2,4=d3,4=d1,3=d2,3=d4,5=2(iv)
dx,x=0,∀x∈X and dx,y=dy,x, ∀x,y∈XThen it is easy to check thatX,d is a complete Branciari metric space which is not a metric space since
(29)4=d1,2>d1,5+d5,2=2.Consider the mappingT:X→X defined by
(30)Tx=3,ifx=1,2,3,5,1,ifx=4.Notice thatT is neither a Θ-contraction nor an L-contraction. Indeed, for x=3 and y=4, we have
(31)d3,4=2=dT3,T4>0.So, in view ofζ2∗ and for any ζ∈L and Θ∈Ω1,2,4, we obtain
(32)Θ2=ΘdT3,T4≤Θd3,4k<Θd3,4=Θ2,and
(33)ζΘdT3,T4,Θd3,4<Θd3,4ΘdT3,T4=1.Therefore, Theorems4 and 9 cannot be used here, but Theorem 15 is applicable. In fact, we have
(34)dTx,Ty=2,ifx=4andy≠4orx≠4andy=4,0,otherwise,and hence, for all x,y∈X,
(35)dTx,Ty>0⇔x=4andy≠4orx≠4and&y=4.Now, we show thatT is a generalized L-contraction with respect to ζk, where ζkt,s=sk/t, for all t,s≥1, L=2, and k=1/2.Define a functionΘ:0,∞⟶1,∞ by
(36)Θt=et.For allx,y∈X with dTx,Ty>0, we have two cases:Case 1.
Ifx=4andy≠1or4 or x≠1or4andy=4, then we have
(37)dx,y=2=dTx,Ty,Nx,y≥1.
Hence,(38)ζΘdTx,Ty,Θdx,y+LNx,y=Θdx,y+LNx,ykΘdTx,Ty≥e1/22+2e2=1.Case 2.
Ifx=4andy=1 or x=1andy=4, then we have
(39)dx,y=4,dTx,Ty=2,Nx,y=0.
Thus,(40)ζΘdTx,Ty,Θdx,y+LNx,y=Θdx,y+LNx,ykΘdTx,Ty=e1/24+0e2=1.Therefore, all the hypotheses of Theorem15 are satisfied, and hence T has a unique fixed point (namely x=3).TakingL=0 in the contractive condition (5), Theorem 15 reduces to the following fixed-point result.Corollary 17 [8].
LetX,d be a complete Branciari metric space and T:X⟶X. If T is an L-contraction with respect to ζ∈L, then T has a unique fixed point.Next, we present the following results, which seem new to the existing literature.Corollary 18.
LetX,d be a complete Branciari metric space and let T:X⟶X be continuous mapping. Suppose that there exist Θ∈Ω1,2,4, L≥0 and k∈0,1 such that (for all x,y∈X)
(41)dTx,Ty>0⇒ΘdTx,Ty≤Θdx,y+LNx,yk,where Nx,y=mindx,Ty,dy,Tx,dx,Tx+dy,Ty/2. Then T has a unique fixed point.Proof.
Observe thatT is a generalized L-contraction with respect to ζkt,s=sk/t. Then, the result follows immediately from Theorem 15.Corollary 19.
LetX,d be a Branciari metric space and let T:X⟶X be continuous mapping such that (for all x,y∈X)
(42)dTx,Ty>0⇒dTx,Ty≤dx,y+LNx,y−φdx,y+LNx,y,where Nx,y=mindx,Ty,d(y,Tx),dx,Tx+dy,Ty/2 and φ:0,∞⟶0,∞ is nondecreasing and lower semicontinuous such that φ−10=0. Then T has a unique fixed point.Proof.
LetΘt=et, for all t>0. From (42), we have
(43)ΘdTx,Ty=edTx,Ty≤edx,y+LNx,y−φdx,y+LNx,y=Θdx,y+LNx,yeφdx,y+LNx,y,for all x,y∈X with dTx,Ty>0.Now, defineφt=lnψΘt, for all t>0, where ψ:1,∞⟶1,∞ is nondecreasing and lower semicontinuous such that ψ−11=1.From (43), we have
(44)ΘdTx,Ty≤Θdx,y+LNx,yψΘdx,y+LNx,y.Takingζt,s=s/tψs and using (44), we have
(45)1≤Θdx,y+LNx,yΘdTx,TyψΘdx,y+LNx,y=ζΘdTx,Ty,Θdx,y+LNx,y.Therefore, all the requirements of Theorem15 are satisfied and hence T has a unique fixed point.


## 4. Fixed-Point Results forL∗-Contractions

Before presenting our main result of this section, we give the following definition.Definition 20.
LetX,d be a Branciari metric space and T:X⟶X. Then T is said to be an L∗-contraction with respect to ζ∈L if there exists Θ∈Ω1,2,4 such that (for all x,y∈X)
(46)dTx,Ty>0⇒ζΘdTx,Ty,ΘMx,y≥1,where Mx,y=maxdx,y,dx,Tx,dy,Ty.Now, we are ready to state and prove the main result of this section.Theorem 21.
LetX,d be a complete Branciari metric space and T:X⟶X an L∗-contraction with respect to ζ∈L. Then T has a unique fixed point.Proof.
Letx0 be an arbitrary point of X. Define a sequence xn⊆X by
(47)xn=Txn−1,∀n∈ℕ.
If there existsn0∈ℕ such that xn0=xn0+1, then xn0 is a fixed point of T and hence the proof is done. Assume that xn≠xn+1, for all n∈ℕ.Using the contractive condition (46) and ζ2∗, we have
(48)1≤ζΘdTxn−1,Txn,ΘMxn−1,xn=ζΘdxn,xn+1,ΘMxn−1,xn<ΘMxn−1,xnΘdxn,xn+1.Consequently, we obtain that(49)Θdxn,xn+1<ΘMxn−1,xn,∀n∈ℕ,where
(50)Mxn−1,xn=maxdxn−1,xn,dxn−1,Txn−1,dxn,Txn=maxdxn−1,xn,dxn,xn+1.IfMxn−1,xn=dxn,xn+1, then inequality (49) becomes
(51)Θdxn,xn+1<Θdxn,xn+1,∀n∈ℕ,which is a contradiction. Hence, we must have Mxn−1,xn=dxn−1,xn, for all n∈ℕ. Therefore, inequality (49) becomes
(52)Θdxn,xn+1<Θdxn−1,xn,∀n∈ℕ,which implies from Θ1 that
(53)dxn,xn+1<dxn−1,xn,∀n∈ℕ.Thus, the sequencedxn−1,xn is decreasing and bounded below by 0, so there exists r≥0 such that limn→∞dxn−1,xn=r. Suppose that r≠0, then it follows from Θ2 that
(54)limn→∞Θdxn−1,xn>1.Takingtn=Θdxn,xn+1 and sn=Θdxn−1,xn, for all n∈ℕ. It is clear from (52), (54), and Θ4 that tn<sn,∀n∈ℕ and limn→∞tn=limn→∞sn>1. Hence, using ζ3∗ we get
(55)1≤limsupn→∞ζtn,sn<1,which is a contradiction. Therefore, r=0, i.e., we have
(56)limn→∞dxn−1,xn=0,∀n∈ℕ.Now, let us assume thatxm=xn, for some m>n. Then, we have xm+1=xn+1. Using (52), we get
(57)Θdxm,xm+1<Θdxm−1,xm<Θdxm−2,xm−1<⋯<Θdxn,xn+1=Θdxm,xm+1,which is a contradiction. This concludes that xm≠xn, for all n≠m.Next, we claim that the sequencexn is a Cauchy sequence in X,d. On the contrary, assume that it is not Cauchy, then we can find two subsequences xnk and xmk of xn such that nk is the smallest index for which
(58)nk>mk>k,dxmk,xnk≥ε,dxmk,xnk−2<ε.By using a similar argument as in the proof of Theorem15, we obtain
(59)limk→∞dxmk,xnk=ε=limk→∞dxmk−1,xnk−1.Now, using (46) and ζ2∗, we have
(60)1≤ζΘdTxmk−1,Txnk−1,ΘMxmk−1,xnk−1=ζΘdxmk,xnk,ΘMxmk−1,xnk−1<ΘMxmk−1,xnk−1Θdxmk,xnk,which implies that
(61)Θdxmk,xnk<ΘMxmk−1,xnk−1,∀k∈ℕ,where Mxmk−1,xnk−1=maxdxmk−1,xnk−1,dxmk−1,xmk,dxnk−1,xnk.Making use of (56), (59), and Remark 11, we deduce that
(62)limk→∞Mxmk−1,xnk−1=maxε,0,0=ε.Now, lettk=Θdxmk,xnk and sk=ΘMxmk−1,xnk−1, for all k∈ℕ. In view of (59), (61), (62), and Θ4, we have tk<sk, for all k∈ℕ and limk→∞tk=limk→∞sk>1. Therefore, using ζ3∗ we obtain
(63)1≤limsupk→∞ζtk,sk<1.This contradiction ensures that the sequencexn is a Cauchy sequence in X,d. As X,d is complete, then there exists u∈X such that
(64)limn→∞dxn,u=0.Without loss of generality, we can assume thatxn≠u and Txn≠Tu, for all n∈ℕ. Suppose that du,Tu>0, it follows from (46) and ζ2∗ that
(65)1≤ζΘdTxn,Tu,ΘMxn,u=ζΘdxn+1,Tu,ΘMxn,u<ΘMxn,uΘdxn+1,Tu,where Mxn,u=maxdxn,u,dxn,xn+1,du,Tu, which implies that
(66)Θdxn+1,Tu<ΘMxn,u.From Remark11i and Lemma 13, we have
(67)limn→∞dxn+1,Tu=limn→∞Mxn,u=du,Tu>0.Lettn=Θdxn+1,Tu and sn=ΘMxn,u, for all n∈ℕ. Then it follows from ζ3∗ that
(68)1≤limsupn→∞ζtn,sn<1,which is a contradiction. Therefore, we conclude that u=Tu, that is, u is a fixed point of T. Finally, we show that the fixed point of the mapping T is unique. Assume that u and z are two distinct fixed points in X. Then du,z=dTu,Tz>0.Using (46) and ζ2∗, we deduce that
(69)1≤ζΘdTu,Tz,ΘMu,z=ζΘdu,z,ΘMu,z<ΘMu,zΘdu,z,where Mu,z=maxdu,z,du,Tu,dz,Tz=du,z, which implies that
(70)Θdu,z<ΘMu,z=Θdu,z,which is a contradiction. Therefore, T has a unique fixed point.Remark 22.
In Theorem15, the mapping T must be continuous, while T in Theorem 21 need not be continuous.To support Theorem21, we give an illustrative example. Precisely, we show that Theorem 21 can be used to cover this example while Theorems 4, 9, and 15 are not applicable.Example 23.
LetX=A∪B, where A=1,2 and B=1/n:n=2,3,4,5. Define a mapping d:X×X⟶0,∞ as follows:
(i)
d1/2,1/3=d1/4,1/5=3/10(ii)
d1/2,1/5=d1/3,1/4=2/10(iii)
d1/2,1/4=d1/5,1/3=6/10(iv)
dx,x=0, dx,y=dy,x, ∀x,y∈B, and(v)
dx,y=x−y if x,y∈A or x∈A, y∈B, or x∈B, y∈A.Observe thatd is not a metric on X, because the triangle inequality is not satisfied on A. To insure this, we have
(71)610=d15,13>d15,14+d14,13=510.It is easy to check thatX,d is a complete Branciari metric space. Let T:X⟶X be defined as
(72)Tx=15,ifx∈1,32,14,ifx∈32,2∪B.SinceT is not continuous at x=3/2, then by Remarks 10 and 5, T is neither a Θ-contraction nor an L-contraction, and hence Theorems 4 and 9 cannot be applied here.Observe thatT is an L∗-contraction with respect to ζ:1,∞×1,∞⟶ℝ, where
(73)ζkt,s=skt,∀t,s∈1,∞,foranyk∈38,1,and Θ:0,∞⟶1,∞, such that Θt=et,∀t∈0,∞.Indeed, forx∈1,3/2 and y∈3/2,2∪B, we have
(74)dTx,Ty=d14,15=310>0,and
(75)ζΘdTx,Ty,ΘMx,y=ΘMx,ykΘdTx,Ty≥e4k/5e3/10=e1/54k−3/2≥1,foranyk∈38,1.Hence, all the hypotheses of Theorem21 are satisfied, and the unique fixed point of T is u=1/4.Notice that due to Remark22, Theorem 15 cannot be applied here.The corollaries that follow are deduced as consequences of Theorem21.Corollary 24.
LetX,d be a complete Branciari metric space and T:X⟶X. Suppose that there exist Θ∈Ω1,2,4 and k∈0,1 such that (for all x,y∈X)
(76)dTx,Ty>0⇒ΘdTx,Ty≤ΘMx,yk,where Mx,y=maxdx,y,dx,Tx,dy,Ty. Then T has a unique fixed point.Proof.
Observe thatT is an L∗-contraction with respect to ζkt,s=sk/t. Then, the result follows immediately from Theorem 21.Remark 25.
Corollary25 is a generalization of Theorem 4 in [17] without assuming condition Θ3.Corollary 26.
LetX,d be a Branciari metric space and T:X⟶X. Assume that (for all x,y∈X)
(77)dTx,Ty>0⇒dTx,Ty≤Mx,y−φMx,y,where Mx,y=maxdx,y,dx,Tx,dy,Ty and φ:0,∞⟶0,∞ is nondecreasing and lower semicontinuous such that φ−10=0. Then T has a unique fixed point.Proof.
LetΘt=et, for all t>0. From (77), we have
(78)ΘdTx,Ty=edTx,Ty≤eMx,y−φMx,y=ΘMx,yeφMx,y,for all x,y∈X with dTx,Ty>0.Now, defineφt=lnψΘt, for all t>0, where ψ:1,∞⟶1,∞ is nondecreasing and lower semicontinuous such that ψ−11=1.From (78), we have
(79)ΘdTx,Ty≤ΘMx,yψΘMx,y.Takingζt,s=s/tψs and using (79), we have
(80)1≤ΘMx,yΘdTx,TyψΘMx,y=ζΘdTx,Ty,ΘMx,y.Therefore, all the requirements of Theorem21 are satisfied, and hence T has a unique fixed point.


## 5. An Application to Fourth-Order Differential Equation

In this section, we discuss application of the fixed-point theorems obtained in the previous sections in solving the following boundary value problem of a fourth-order differential equation:(81)x′′′′t=ft,xt,x′t,x′′t,x′′′t,t∈0,1,x0=x′0=x′′1=x′′′1=0,where f:0,1×ℝ4⟶ℝ is a continuous function. Let X=C0,1 be the space of all continuous functions defined on 0,1. Define a metric d:X×X⟶ℝ by
(82)dx,y=maxt∈0,1xt−yt,forallx,y∈X.It is known thatX,d is a complete Branciari distance space. The green function associated to (81) is defined by
(83)Gs,t=16t23s−t,0≤t≤s≤1,16s23t−s,0≤s≤t≤1.Now, we prove the following result on the solution of the boundary value problem (81).Theorem 27.
Assume that the following conditions are satisfied:(i)
f:0,1×ℝ4⟶ℝ is a continuous function(ii)
There existsτ>0 such that, for all x,y∈X and s∈0,1(84)fs,x,x′−fs,y,y′≤8e−τmaxxs−ys,xs−Txs,ys−Tys,where T:X⟶X is defined by
(85)Txt=∫01Gt,sfs,xs,x′sds.Then (81) has a unique solution in X.Proof.
Observe thatx∈X is a solution of (81) if and only if x∈X is a solution of the integral equation
(86)xt=∫01Gt,sfs,xs,x′sds,∀x∈X.In view of conditionii. For all x,y∈X with dTx,Ty>0 and for all t∈0,1, we have
(87)Txt−Tyt=∫01Gt,sfs,xs,x′sds−∫01Gt,sfs,ys,y′sds≤∫01Gt,sfs,xs,x′s−fs,ys,y′sds≤8e−τ∫01Gt,smaxx−y,x−Tx,y−Tyds≤8e−τMx,ysupt∈0,1∫01Gt,sds,where Mx,y=maxdx,y,dx,Tx,dy,Ty. As ∫01Gt,sds=t4/24−t3/6+t2/4, for all t∈0,1, supt∈0,1∫01Gt,sds=1/8, we obtain
(88)dTx,Ty≤e−τMx,y,or
(89)edTx,Ty≤eMx,ye−τ.Observe thate−τ∈0,1 as τ>0. Therefore, for all x,y∈X, we obtain
(90)ζΘdTx,Ty,ΘMx,y=ΘMx,ykΘdTx,Ty=eMx,ye−τedTx,Ty≥1,where Θt=et, ζt,s=sk/t, and k=e−τ. Thus, all the hypotheses of Theorem 21 are satisfied, and hence T has a unique fixed point in X which is a solution of (81).Theorem 28.
Assume that the following conditions are satisfied:(i)
f:0,1×ℝ4⟶ℝ is a continuous function(ii)
There existsτ>0 and λ≥0 such that for all x,y∈X and s∈0,1(91)fs,x,x′−fs,y,y′≤8e−τxs−ys+λNx,y,where
(92)Nx,y=mindx,Ty,dy,Tx,dx,Tx+dy,Ty2,and T:X⟶X is defined by
(93)Txt=∫01Gt,sfs,xs,x′sds.Then (81) has a unique solution in X.Proof.
The proof can be done using similar arguments of the proof of Theorem28 and applying Theorem 15.



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