# Several Fixed-Point Theorems forF-Contractions in Complete Branciari b-Metric Spaces and Applications

Authors: Lili Chen; Shuai Huang; Chaobo Li; Yanfeng Zhao
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we prove the existence and uniqueness of fixed points forF-contractions in complete Branciari b-metric spaces. Furthermore, an example for supporting the related result is shown. We also present the concept of the weak well-posedness of the fixed-point problem of the mapping T and discuss the weak well-posedness of the fixed-point problem of an F-contraction in complete Branciari b-metric spaces. Besides, we investigate the problem of common fixed points for F-contractions in above spaces. As an application, we apply our main results to solving the existence and uniqueness of solutions for a class of the integral equation and the dynamic programming problem, respectively.


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## Body

## 1. Introduction

In the last decades, one can observe a huge amount of interest for the development of fixed-point theory, since it has a lot of applications, especially in metric spaces, see [1–6]. In 1993, Czerwik [7] introduced the concepts of b-metric spaces by weakening the coefficient of the triangle inequality and generalized Banach’s Contraction Principle to these spaces. Subsequently, Boriceanu [8] obtained some concrete examples of b-metric spaces and studied the fixed-point properties of set-valued operators in b-metric spaces. In 2000, Branciari [9] suggested a new distance function to improve the notion of metric via substituting the triangle inequality with the quadrilateral inequality and extended Banach’s Contraction Principle to Branciari metric spaces. In 2015, George et al. [10] combined Branciari metric spaces with b-metric spaces to introduce a generalization of metric spaces-rectangular b-metric spaces (or Branciari b-metric spaces). They also proved Banach’s Contraction Principle and presented the generalized form of Kannan’s fixed-point theorem in above spaces. In 2018, Karapinar [11] revisited Kannan’s fixed-point theorem under the aspect of interpolation and proposed a new Kannan-type contraction to maximize the rate of convergence. For more details about fixed-point theorems in metric spaces, please see [12–17].In fact, fixed-point properties of Branciarib-metric spaces have aroused great interest of many researchers who have extended the results in Branciari metric spaces and b-metric spaces to Branciari b-metric spaces. However, sometimes it is difficult to achieve the desired results in Branciari b-spaces; there are still a large number of problems to be solved. Based on the above-mentioned discussions, we discuss the problem of the existence and uniqueness of fixed points for F-contractions in complete Branciari b-metric spaces which extend the results in [18]. What is more, an example for supporting the result is presented. The concept of the weak well-posedness of the fixed-point problem of the mapping T is also introduced, and the weak well-posedness of the fixed-point problem of an F-contraction satisfying some conditions in complete Branciari b-metric spaces is investigated. Besides, the problem of common fixed points for two F-contractions is studied in above spaces. As an application, by using main results, the existence and uniqueness of solutions for a class of the integral equation and the dynamic programming problem are solved, respectively.


## 2. Preliminaries

Firstly, we recall some basic notations and facts that we are going to use later.Definition 1 (see [19]).
LetU be a nonempty set and s≥1 be a real number. A function d:U×U⟶0,∞ is called a b-metric if the following conditions are satisfied, for every x,y,z∈U:
(A1)dx,y=0 if and only if x=y
(A2)dx,y=dy,x
(A3)dx,y≤sdx,z+dz,yIn this case, the pairU,d is called a b-metric space with constant s≥1.Definition 2 (see [10]).
LetU be a nonempty set and s≥1 be a real number. A function d:U×U⟶0,∞ is called a Branciari b-metric if the following conditions are satisfied, for every x,y∈U:
(B1)dx,y=0 if and only if x=y
(B2)dx,y=dy,x
(B3)dx,y≤sdx,u+du,v+dv,y for all distinct points u,v∈U\x,yIn this case, the pairU,d is called a Branciari b-metric space with constant s≥1.Definition 3 (see [18]).
A functionF:0,∞⟶ℝ belongs to F if it satisfies the following conditions:
(F1)F is strictly increasing
(F2) There existsk∈0,1 such that limx→0+xkFx=0In [18], they omitted Wardowski’s (F2) condition from the above definition. Explicitly, they do not require that (WF2) if αnn∈N is a sequence of positive real numbers, then limn→∞αn=0 if and only if limn→∞Fαn=−∞The reason for this is the following lemma.Lemma 4 (see [18]).
IfF:0,∞⟶ℝ is an increasing function and αnn∈N⊂0,+∞ is a decreasing sequence such that limn→∞Fαn=−∞, then limn→∞αn=0.We can also see some properties concerningFs,τ and Fsτ′.Definition 5 (see [19]).
Lets≥1 and τ>0. We say that F∈F belongs to Fs,τ if it also satisfies Fsτ if inf F=−∞ and x,y,z∈0,∞ are such that τ+Fsx≤Fy and τ+Fsy≤Fz, then
(1)τ+Fs2x≤Fsy.In [20], the authors introduced the following condition (F4) in Definition3.1:Fsτ′ if αnn∈N⊂0,∞ is a sequence such that τ+Fsαn≤Fαn−1, for all n∈N and for some τ≥0, then τ+Fsnαn≤Fsn−1αn−1, for all n∈N∗.Proposition 6 (see [18]).
IfF is increasing, then Fsτ is equivalent to Fsτ′.Definition 7 (see [18]).
LetU,d be a b-metric space with constant s≥1 and T:U⟶U be an operator. If there exist τ>0 and F∈Fs,τ, such that for all x,y∈U, the inequality dTx,Ty>0 implies
(2)Fτ+Fs⋅dTx,Ty≤Fdx,y,then, T is called an F-contraction.


## 3. Main Results

In this section, we mainly focus on the existence and uniqueness of fixed points forF-contractions, the weak well-posedness of the fixed-point problem of an F-contraction, and criteria for common fixed points of two F-contractions in complete Branciari b-metric space. Moreover, the existence of fixed points for Hardy-Rogers-type set-valued F-contractions in above spaces is shown. In the sequel, we denote by ℝ the set of real numbers and by ℕ the set of positive integer numbers.


### 3.1. Existence and Uniqueness of Fixed Points forF-Contractions

Theorem 8.
LetU,d be a complete Branciari b-metric space with constant s≥1 and T:U⟶U be an F-contraction for F∈Fs,τ. Then, T has a unique fixed point x∗. Furthermore, for any x0∈U, the sequence xn satisfying xn=Txn−1 is convergent.Proof.
Firstly, we show thatT has at most one fixed point. Suppose that x∗ and y∗ are two different fixed points of T. That is, Tx∗=x∗≠y∗=Ty∗. It follows that
(3)dTx∗,Ty∗=dx∗,y∗>0.
Therefore, we can applyF to get
(4)τ+Fs⋅dTx∗,Ty∗≤Fdx∗,y∗≤Fs⋅dx∗,y∗=Fs⋅dTx∗,Ty∗,which implies τ≤0, a contradiction. Hence, T has at most one fixed point.
Next, we prove the existence of fixed points ofT. For any x0∈U, set xn=Txn−1, γn=dxn+1,xn, and βn=dxn+2,xn with γ0=dx1,x0 and β0=dx2,x0. Now, we consider the following two cases:
(i)
If there existsn0∈ℕ such that xn0=xn0−1, then we have Txn0−1=xn0−1. This shows x∗=xn0−1, a fixed point of T. Therefore, the proof is finished(ii)
Ifxn≠xn−1, for any n∈ℕ, then we have γn>0, for each n∈ℕ. Hence, F implies Fsγn+1≤Fγn−τ, for every n∈ℕ
By Proposition6, we obtain
(5)Fsn+1γn+1≤Fsnγn−τ,∀n∈ℕ.
Furthermore, for anyn∈ℕ, we have
(6)Fsnγn≤Fsn−1γn−1−τ≤Fsn−2γn−2−2τ≤⋯≤Fγ0−nτ.
Sincelimn→∞Fγ0−nτ=−∞, then the inequality (6) implies
(7)limn→∞Fsnγn=−∞.
On the other hand, from (5) and (F1), it follows that the sequence snγnn=1∞ is decreasing. Furthermore, we can apply Lemma 4 to get
(8)limn→∞snγn=0.
According to (F2), there existsk∈0,1 such that
(9)limn→∞snγnkFsnγn=0.
Multiplying (6) by snγnk results
(10)0≤nsnγnkτ+snγnkFsnγn≤snγnkFγ0,∀n∈ℕ,which implies limn→∞nsnγnk=0. It follows that there exists n1∈ℕ such that nsnγnk≤1, for all n≥n1. Thus,
(11)snγn≤1n1/k,∀n≥n1,which shows that the series ∑i=1∞siγi is convergent.Now, we prove thatxnn∈ℕ is a Cauchy sequence. For all n,p∈ℕ, we drive the proof into two cases.
(a)
Ifp>2 is odd, we can obtain(12)dxn+3,xn≤sdxn+3,xn+2+sdxn+2,xn+1+sdxn+1,xn,dxn+5,xn≤sdxn+5,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+5,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2+sγn+1+sγn.Hence, we deduce that(13)dxn+p,xn≤sdxn+p,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2+sγn+1+sγn≤s3dxn+p,xn+6+s3γn+5+s3γn+4+s2γn+3+s2γn+2+sγn+1+sγn⋮≤sp−1/2γn+p−1+sp−1/2γn+p−2+sp−1/2γn+p−3+sp−3/2γn+p−4+sp−3/2γn+p−5+⋯+s2γn+2+sγn+1+sγn≤sp+1/2γn+p−1+sp/2γn+p−2+sp−1/2γn+p−3+sp−2/2γn+p−4+sp−3/2γn+p−5+⋯+s3/2γn+1+s2/2γn≤sp+1γn+p−1+spγn+p−2+sp−1γn+p−3+sp−2γn+p−4+sp−3γn+p−5+⋯+s3γn+1+s2γn≤1sn−2sn+p−1γn+p−1+sn+p−2γn+p−2+sn+p−3γn+p−3+⋯+sn+1γn+1+snγn=1sn−2∑i=nn+p−1siγi≤1sn−2∑i=n∞siγi.(b)
Ifp>2 is even, we can get(14)dxn+4,xn≤sdxn+4,xn+2+sdxn+2,xn+1+sdxn+1,xn.Furthermore, we conclude that(15)dxn+p,xn≤sdxn+p,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2≤s3dxn+p,xn+6+s3γn+5+s3γn+4+s2γn+3+s2γn+2⋮≤sp−2/2dxn+p,xn+p−2+sp−2/2γn+p−3+sp−2/2γn+p−4+sp−4/2γn+p−5+sp−4/2γn+p−6+⋯+sγn+1+sγn≤sp−2/2dxn+p,xn+p−2+sp−1/2γn+p−3+sp−2/2γn+p−4+sp−3/2γn+p−5+sp−4/2γn+p−6+⋯+s3/2γn+1+s2/2γn≤sp−2/2dxn+p,xn+p−2+1sn−2sn+p−3γn+p−3+sn+p−4γn+p−4+⋯+sn+1γn+1+snγn≤sp−2/2dxn+p,xn+p−2+1sn−2∑i=nn+p−3siγi≤sp−2/2βn+p−2+1sn−2∑i=n∞siγi.Sinceβn≥0, we can assume that βn>0,∀n∈ℕ. By a similar method, replacing γn with βn in (11), there exists n2∈ℕ such that
(16)snβn≤1n1/k,∀n≥n2,which implies limn→∞snβn=0 and limn→∞βn=0. Together (a) with (b), for every p∈ℕ, letting n⟶∞, we deduce that
(17)dxn+p,xn⟶0.Thus,xnn∈N is a Cauchy sequence. Since U,d is complete, there exists x∗∈U such that limn→∞xn=x∗. On the other hand, we notice that
(18)FdTx,Ty≤τ+FdTx,Ty≤τ+Fs⋅dTx,Ty≤Fdx,yholds for all x,y∈U with dTx,Ty>0. Since F is increasing, then
(19)dTx,Ty≤dx,y.It follows that(20)0≤dxn+1,Tx∗≤dxn,x∗⟶0asn⟶∞.Hence, we getx∗=Tx∗, which completes the proof.Next, we will give an example to support Theorem8.Example 1.
LetU=A∪B, where A=1/n:n∈2,3,4,5 and B=1,2. For any x,y∈U, we define d:U×U⟶0,∞ by
(21)dx,y=x−y2.Then, it is not difficult to see thatU,d is a Branciari b-metric space with constant s=12/5>1.LetT:U⟶U be a mapping satisfying
(22)Tx=14,x∈A,15,x∈B.Now, we verify thatT is an F-contraction satisfying the condition F. We take x∈A, y∈B, and τ=0.1. Then, dTx,Ty=d1/4,1/5=1/400>0 and
(23)0.1+125⋅dTx,Ty=0.106<0.5=dx,y.LetF:0,∞⟶ℝ be a mapping defined by Fx=x, then it is easy to see that F∈Fs,τ. In addition, we can get
(24)τ+Fs⋅dTx,Ty≤Fdx,y.Hence,T satisfies the conditions of Theorem 8 and x=1/4 is the unique fixed point of T.


### 3.2. Weak Well-Posedness for the Fixed-Point Problem of anF-Contraction

The notion of well-posedness of a fixed-point problem has evoked much interest to several researchers [21–23]. Now, we introduce the concept of weak well-posedness for the fixed-point problem of the mapping T.Definition 9.
LetU,d be a metric space and T:U⟶U be a mapping. The fixed-point problem of T is said to be weakly well-posed if
(1)
T has a unique fixed point x∗ in U(2)
For any sequencexnn∈ℕ in U with limn→∞dxn,Txn=0 andlimn,m→∞dTxn,Txm=0, we have limn→∞dxn,x∗=0Theorem 10.
LetU,d be a complete Branciari b-metric space with constant s≥1 and T:U⟶U be an F-contraction for F∈Fs,τ with the continuous differentiable function F′ and limx→∞F′x<∞. Then, the fixed-point problem of T is weakly well-posed.Proof.
From Theorem8, it follows that T satisfies (5). Let x∗ be the unique fixed point of T. For any sequence xnn∈ℕ in U with limn→∞dxn,Txn=0 and limn,m→∞dTxn,Txm=0, in general, the sequence xnn∈ℕ can be decomposed into two subsequences xn=ynkk∈ℕ∪znkk∈ℕ, where Tynk=x∗ and Tznk≠x∗ for any k∈ℕ. Consequently, we can prove the following two assertions: Fdxn,x∗−FsdTxn,Tx∗≥τ.
(W1) ifxnn∈ℕ is such that Txn=x∗ for all n∈ℕ, then limn→∞dxn,x∗=0
(W2) ifxnn∈ℕ is such that Txn≠x∗ for all n∈ℕ, then limn→∞dxn,x∗=0In fact, (W1) is trivial, sincedxn,x∗=dxn,Txn⟶0 as n⟶∞. Now, we prove (W2). Because of the uniqueness of the fixed point of T, we can assume that there exists n0∈ℕ such that Txn≠xn for n≥n0. If Txn≠x∗ for any n∈ℕ, then we can apply F to get
(25)Fdxn,x∗−FsdTxn,Tx∗≥τ.SinceF is strictly increasing, we have dxn,x∗>sdTxn,Tx∗. Then,
(26)Fdxn,x∗−FsdTxn,Tx∗dxn,x∗−sdTxn,Tx∗≥τdxn,x∗−sdTxn,Tx∗.Then there exists a constantcnbetween dxn,x∗ and sdTxn,Tx∗ such that
(27)F′cn=Fdxn,x∗−FsdTxn,Tx∗dxn,x∗−sdTxn,Tx∗≥τdxn,x∗−sdTxn,Tx∗,which implies
(28)0<τF′cn≤dxn,x∗−sdTxn,Tx∗.Actually, we can assume that there exists a distinct subsequenceTxnk of {Txn}. Otherwise, there exists x0∈U and n1∈ℕ such that Txn=x0 for n≥n1. Since limn→∞dxn,Txn=0, we get limn→∞dxn,x0=0. If x0≠x∗, then x0≠Tx0 because of the uniqueness of the fixed point of T. For n≥n1, we obtain x0=Txn≠Tx0. Since T is an F-contraction, together with (19), we have
(29)0<dTxn,Tx0≤dxn,x0⟶0asn⟶∞,which is a contradiction. Hence, there exist m,q,n>n0 (m>q>n) such that Txm≠Txq≠Txn≠xn. Then,
(30)dxn,x∗≤sdxn,Txm+dTxm,Txn+dTxn,x∗.Combining with (28), we deduce that
(31)0<τF′cn≤sdxn,Txm+dTxm,Txn.Since(32)0≤dxn,Txm≤sdxn,Txn+dTxn,Txq+dTxq,Txm⟶0asn⟶∞.Then,(33)0<τF′cn≤sdxn,Txm+dTxm,Txn⟶0asn⟶∞.Therefore, we getlimn→∞F′cn=∞ which implies limn→∞cn=0 since F′ is continuous and limx→∞F′x<∞. Hence, we deduce that limn→∞sdTxn,Tx∗=0. From (30), it follows that limn→∞dxn,x∗=0.


### 3.3. The Problem of Common Fixed Points for TwoF-Contractions

Now, we continue to solve the problem of common fixed points for twoF-contraction mappings in a complete Branciari b-metric space.Theorem 11.
LetU,d be a complete Branciari b-metric space with constant s>1. If there exist τ>0 and F∈Fs,τ, such that S,T:U⟶U are two F-contraction mappings on U and satisfy
(34)τ+FdSx,Ty≤Fadx,y+bdx,Sx+cdy,Ty,for any a,b,c∈0,1 with a+b+c<1, sb<1, and mindSx,Ty,dx,y,dx,Sx,dy,Ty>0 for any x,y∈U×U. Then, there exists a unique common fixed point of the pair mappings S,T.Proof.
Letx0∈U, we define xn⊆U by
(35)Sx2j=x2j+1,Tx2j+1=x2j+2,j=0,1,2,⋯.
Combining with (34) and (35), we have
(36)τ+Fdx2j+1,x2j+2=τ+FdSx2j,Tx2j+1≤Fadx2j,x2j+1+bdx2j,Sx2j+cdx2j+1,Tx2j+1=Fadx2j,x2j+1+bdx2j,x2j+1+cdx2j+1,x2j+2.
Letλ=a+b/1−c, it is not difficult to see 0<λ<1 since a+b+c<1. Thanks to the strictly monotone increasing property of F, we deduce
(37)dx2j+1,x2j+2<a+b1−cdx2j,x2j+1=λdx2j,x2j+1.
Similarly,(38)dx2j+2,x2j+3<a+b1−cdx2j+1,x2j+2=λdx2j+1,x2j+2.
Hence,(39)dxn,xn+1<λdxn−1,xn,n∈ℕ.
For anyn∈ℕ, we obtain
(40)dxn,xn+1<λdxn−1,xn<λ2dxn−2,xn−1<⋯<λndx0,x1.
Notice that(41)τ+Fsdx1,x3=τ+FsdSx0,Sx2≤Fdx0,x2,τ+Fsdx2,x4=τ+FsdTx1,Tx3≤Fdx1,x3.
SinceF is strictly monotone increasing, we have
(42)dx1,x3≤1sdx0,x2,dx2,x4≤1sdx1,x3≤1s2dx0,x2.
By induction, we get(43)dxn,xn+2<1sndx0,x2,n∈ℕ.Next, we will show thatxn is a Cauchy sequence in U. We consider the following two cases.
(i)
Letm=n+p, if p is odd and p>2, we have(44)dxn,xm≤sdxn,xn+1+dxn+1,xn+2+dxn+2,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s2dxn+4,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s3dxn+4,xn+5+s3dxn+5,xn+6+⋯+sm−n/2dxm−2,xm−1+sm−n/2dxm−1,xm≤sλndx0,x1+sλn+1dx0,x1+s2λn+2dx0,x1+s2λn+3dx0,x1+s3λn+3dx0,x1+s3λn+4dx0,x1+⋯+sm−n/2λm−2dx0,x1+sm−n/2λm−1dx0,x1≤sλn+s2λn+2+s3λn+4+⋯+sm−n/2λm−2dx0,x1+sλn+1+s2λn+3+s3λn+5+⋯+sm−n/2λm−1dx0,x1≤sλn+s2λn+2+s3λn+4+⋯+sm−n/2λm−21+λdx0,x1≤sλn1+sλ2+s2λ4+s3λ6+⋯+sm−n−2/2λm−n−21+λdx0,x1≤sλn⋅1−sm−n/2λm−n1−sλ21+λdx0,x1≤sλn⋅1−sp/2λp1−sλ21+λdx0,x1.(ii)
Letm=n+p, if p is even and p>2, we have(45)dxn,xm≤sdxn,xn+1+dxn+1,xn+2+dxn+2,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s2dxn+4,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s3dxn+4,xn+5+s3dxn+5,xn+6+⋯+sm−n−2/2dxm−4,xm−3+sm−n−2/2dxm−3,xm−2+sm−n−2/2dxm−2,xm≤sλndx0,x1+sλn+1dx0,x1+s2λn+2dx0,x1+s2λn+3dx0,x1+s3λn+3dx0,x1+s3λn+4dx0,x1+⋯+sm−n−2/2λm−4dx0,x1+sm−n−2/2λm−3dx0,x1+sm−n−2/2dxm−2,xm≤sλn+s2λn+2+s3λn+4+⋯+sm−n−2/2λm−4dx0,x1+sλn+1+s2λn+3+s3λn+5+⋯+sm−n−2/2λm−3dx0,x1+sm−n−2/2dxm−2,xm≤sλn+s2λn+2+s3λn+4+⋯+sm−n−2/2λm−41+λdx0,x1+sm−n−2/2dxm−2,xm≤sλn1+sλ2+s2λ4+s3λ6+⋯+sm−n−4/2λm−n−41+λdx0,x1+sm−n−2dxm−2,xm≤sλn⋅1−sm−n−2/2λm−n−21−sλ21+λdx0,x1+s−ndx0,x2≤sλn⋅1−sp−2/2λp−21−sλ21+λdx0,x1+s−ndx0,x2.Lettingm,n⟶∞, we obtain that dxn,xm⟶0 for all p>2. Hence, xn is a Cauchy sequence in U. From the completeness of U, it follows that there exists z∗∈U such that
(46)limn→∞dxn,z∗=0.Next, we verify thatz∗ is a fixed point of the mapping S. Indeed, if dSz∗,z∗>0, then
(47)τ+FdSz∗,x2j+2≤Fadz∗,x2j+1+bdz∗,Sz∗+cdx2j+1,x2j+2.By using the strictly monotone increasing property ofF, we get
(48)dSz∗,x2j+2<adz∗,x2j+1+bdz∗,Sz∗+cdx2j+1,x2j+2.We can also see that
(49)dSz∗,z∗<sdSz∗,x2j+2+dx2j+2,x2j+1+dx2j+1,z∗.It follows that(50)1sdSz∗,z∗≤liminfj→∞dSz∗,x2j+2≤limsupj→∞dSz∗,x2j+2≤bdz∗,Sz∗.Hence, we deduce that1/s≤b which contradicts the fact sb<1. Therefore, we obtain dSz∗,z∗=0, i.e., Sz∗=z∗. Similarly, we can get Tz∗=z∗. Therefore, we have
(51)Tz∗=Sz∗=z∗.Next, we will show thatS and T have a unique common fixed point. Suppose that z∗, z∗∗ are two distinct common fixed points of S and T. Then,
(52)τ+Fdz∗,z∗∗=τ+FdSz∗,Tz∗∗≤Fadz∗,z∗∗+bdz∗,Sz∗+cdz∗∗,Tz∗∗=Fadz∗,z∗∗+bdz∗,z∗+cdz∗∗,z∗∗.In view of the strictly monotone increasing property ofF, we obtain 1−adz∗,z∗∗<0, which is a contradiction. It means z∗=z∗∗.Next, we discuss the existence of fixed points for Hardy-Rogers-type set-valuedF-contractions in a complete Branciari b-metric space.Lemma 12 (see [24]).
IfF is right continuous and satisfies (F1), then
(53)FinfA=infFA,for all A∈0,∞ with infA>0.Definition 13 (see [24]).
LetU,d be a Branciari b-metric space and CBU be the family of all nonempty closed and bounded subsets of U. Let H:CBU×CBU⟶0,∞ be a function defined by HA,B=maxsupx∈Adx,B,supy∈Bdy,A for all A,B∈CBU, where dx,B=infdx,y:y∈B. Then, H defines a metric on CBU called the Hausdorff metric induced by d.Theorem 14.
LetU,d be a complete Branciari b-metric space with constant s≥1. Assume that there exist τ>0 and F∈Fs,τ such that T:U⟶CBU satisfies
(54)2τ+FHTx,Ty≤Fadx,y+bdx,Tx+cdy,Ty,for all x,y∈U and a,b,c≥0 with a+b+c<1 and sc<1. If F is right continuous, then T has a fixed point in U.Proof.
Letx0 be an arbitrary point of U and choose x1∈Tx0. If x1∈Tx1, then x1 is a fixed point of T and the proof is finished. Assume that x1∉Tx1, then Tx0≠Tx1. Since F is right continuous and Tx1 is closed, it follows from Lemma 12 that
(55)Fdx1,Tx1=infy∈Tx1Fdx1,y.
Noticing thatdx1,Tx1≤HTx0,Tx1, then there exists x2∈Tx1 such that
(56)Fdx1,x2−τ<Fdx1,Tx1≤FHTx0,Tx1.
Consequently, we get(57)Fdx1,x2<FHTx0,Tx1+τ,which implies
(58)2τ+Fdx1,x2<2τ+FHTx0,Tx1+τ≤Fadx0,x1+bdx0,Tx0+cdx1,Tx1+τ≤Fadx0,x1+bdx0,x1+cdx1,x2+τ=Fa+bdx0,x1+cdx1,x2+τ.
SinceF is strictly increasing, we deduce
(59)dx1,x2<a+bdx0,x1+cdx1,x2,and hence,
(60)1−cdx1,x2<a+bdx0,x1.
Sincea+b+c<1, we obtain
(61)dx1,x2<a+b1−cdx0,x1<dx0,x1.
Combining with (58), we get
(62)τ+Fdx1,x2<Fdx0,x1.
By induction, we can get a sequencexn⊂U satisfying dxn,xn+1 is decreasing, xn∉Txn, xn+1∈Txn, and τ+Fdxn,xn+1<Fdxn−1,xn for all n∈ℕ.
Letγn=dxn,xn+1 and βn=dxn,xn+2. Then, we have
(63)Fγn+1<Fγn−τ,for all n∈ℕ. Similarly to the proof of Theorem 8 (while s=1), we deduce that xn is a Cauchy sequence. Since U is complete, there exists x∗∈U such that xn⟶x∗ as n⟶∞. Now, we prove that x∗ is a fixed point of T. Since
(64)2τ+Fdxn,Tx∗≤2τ+FHTxn−1,Tx∗≤Fadxn−1,x∗+bdxn−1,Txn−1+cdx∗,Tx∗,dxn−1,Txn−1≤dxn−1,xn,we deduce that
(65)dxn,Tx∗<adxn−1,x∗+bdxn−1,xn+cdx∗,Tx∗.
It follows that(66)limsupn→∞dxn,Tx∗≤cdx∗,Tx∗.
Noting that(67)dx∗,Tx∗=infy∈Tx∗dx∗,y≤sinfy∈Tx∗dx∗,xn+1+dxn+1,xn+dxn,y≤sdx∗,xn+1+dxn+1,xn+sdxn,Tx∗,we obtain
(68)dx∗,Tx∗≤slimsupn→∞dxn,Tx∗.
Combining with (66) and (68), we get
(69)limsupn→∞dxn,Tx∗≤sclimsupn→∞dxn,Tx∗.
Sincesc<1, then limsupn→∞dxn,Tx∗=0 which implies dx∗,Tx∗=0. Hence, x∗ is a fixed point of T since Tx∗ is closed.


## 3.1. Existence and Uniqueness of Fixed Points forF-Contractions

Theorem 8.
LetU,d be a complete Branciari b-metric space with constant s≥1 and T:U⟶U be an F-contraction for F∈Fs,τ. Then, T has a unique fixed point x∗. Furthermore, for any x0∈U, the sequence xn satisfying xn=Txn−1 is convergent.Proof.
Firstly, we show thatT has at most one fixed point. Suppose that x∗ and y∗ are two different fixed points of T. That is, Tx∗=x∗≠y∗=Ty∗. It follows that
(3)dTx∗,Ty∗=dx∗,y∗>0.
Therefore, we can applyF to get
(4)τ+Fs⋅dTx∗,Ty∗≤Fdx∗,y∗≤Fs⋅dx∗,y∗=Fs⋅dTx∗,Ty∗,which implies τ≤0, a contradiction. Hence, T has at most one fixed point.
Next, we prove the existence of fixed points ofT. For any x0∈U, set xn=Txn−1, γn=dxn+1,xn, and βn=dxn+2,xn with γ0=dx1,x0 and β0=dx2,x0. Now, we consider the following two cases:
(i)
If there existsn0∈ℕ such that xn0=xn0−1, then we have Txn0−1=xn0−1. This shows x∗=xn0−1, a fixed point of T. Therefore, the proof is finished(ii)
Ifxn≠xn−1, for any n∈ℕ, then we have γn>0, for each n∈ℕ. Hence, F implies Fsγn+1≤Fγn−τ, for every n∈ℕ
By Proposition6, we obtain
(5)Fsn+1γn+1≤Fsnγn−τ,∀n∈ℕ.
Furthermore, for anyn∈ℕ, we have
(6)Fsnγn≤Fsn−1γn−1−τ≤Fsn−2γn−2−2τ≤⋯≤Fγ0−nτ.
Sincelimn→∞Fγ0−nτ=−∞, then the inequality (6) implies
(7)limn→∞Fsnγn=−∞.
On the other hand, from (5) and (F1), it follows that the sequence snγnn=1∞ is decreasing. Furthermore, we can apply Lemma 4 to get
(8)limn→∞snγn=0.
According to (F2), there existsk∈0,1 such that
(9)limn→∞snγnkFsnγn=0.
Multiplying (6) by snγnk results
(10)0≤nsnγnkτ+snγnkFsnγn≤snγnkFγ0,∀n∈ℕ,which implies limn→∞nsnγnk=0. It follows that there exists n1∈ℕ such that nsnγnk≤1, for all n≥n1. Thus,
(11)snγn≤1n1/k,∀n≥n1,which shows that the series ∑i=1∞siγi is convergent.Now, we prove thatxnn∈ℕ is a Cauchy sequence. For all n,p∈ℕ, we drive the proof into two cases.
(a)
Ifp>2 is odd, we can obtain(12)dxn+3,xn≤sdxn+3,xn+2+sdxn+2,xn+1+sdxn+1,xn,dxn+5,xn≤sdxn+5,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+5,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2+sγn+1+sγn.Hence, we deduce that(13)dxn+p,xn≤sdxn+p,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2+sγn+1+sγn≤s3dxn+p,xn+6+s3γn+5+s3γn+4+s2γn+3+s2γn+2+sγn+1+sγn⋮≤sp−1/2γn+p−1+sp−1/2γn+p−2+sp−1/2γn+p−3+sp−3/2γn+p−4+sp−3/2γn+p−5+⋯+s2γn+2+sγn+1+sγn≤sp+1/2γn+p−1+sp/2γn+p−2+sp−1/2γn+p−3+sp−2/2γn+p−4+sp−3/2γn+p−5+⋯+s3/2γn+1+s2/2γn≤sp+1γn+p−1+spγn+p−2+sp−1γn+p−3+sp−2γn+p−4+sp−3γn+p−5+⋯+s3γn+1+s2γn≤1sn−2sn+p−1γn+p−1+sn+p−2γn+p−2+sn+p−3γn+p−3+⋯+sn+1γn+1+snγn=1sn−2∑i=nn+p−1siγi≤1sn−2∑i=n∞siγi.(b)
Ifp>2 is even, we can get(14)dxn+4,xn≤sdxn+4,xn+2+sdxn+2,xn+1+sdxn+1,xn.Furthermore, we conclude that(15)dxn+p,xn≤sdxn+p,xn+2+sdxn+2,xn+1+sdxn+1,xn≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2≤s2dxn+p,xn+4+s2dxn+4,xn+3+s2dxn+3,xn+2≤s3dxn+p,xn+6+s3γn+5+s3γn+4+s2γn+3+s2γn+2⋮≤sp−2/2dxn+p,xn+p−2+sp−2/2γn+p−3+sp−2/2γn+p−4+sp−4/2γn+p−5+sp−4/2γn+p−6+⋯+sγn+1+sγn≤sp−2/2dxn+p,xn+p−2+sp−1/2γn+p−3+sp−2/2γn+p−4+sp−3/2γn+p−5+sp−4/2γn+p−6+⋯+s3/2γn+1+s2/2γn≤sp−2/2dxn+p,xn+p−2+1sn−2sn+p−3γn+p−3+sn+p−4γn+p−4+⋯+sn+1γn+1+snγn≤sp−2/2dxn+p,xn+p−2+1sn−2∑i=nn+p−3siγi≤sp−2/2βn+p−2+1sn−2∑i=n∞siγi.Sinceβn≥0, we can assume that βn>0,∀n∈ℕ. By a similar method, replacing γn with βn in (11), there exists n2∈ℕ such that
(16)snβn≤1n1/k,∀n≥n2,which implies limn→∞snβn=0 and limn→∞βn=0. Together (a) with (b), for every p∈ℕ, letting n⟶∞, we deduce that
(17)dxn+p,xn⟶0.Thus,xnn∈N is a Cauchy sequence. Since U,d is complete, there exists x∗∈U such that limn→∞xn=x∗. On the other hand, we notice that
(18)FdTx,Ty≤τ+FdTx,Ty≤τ+Fs⋅dTx,Ty≤Fdx,yholds for all x,y∈U with dTx,Ty>0. Since F is increasing, then
(19)dTx,Ty≤dx,y.It follows that(20)0≤dxn+1,Tx∗≤dxn,x∗⟶0asn⟶∞.Hence, we getx∗=Tx∗, which completes the proof.Next, we will give an example to support Theorem8.Example 1.
LetU=A∪B, where A=1/n:n∈2,3,4,5 and B=1,2. For any x,y∈U, we define d:U×U⟶0,∞ by
(21)dx,y=x−y2.Then, it is not difficult to see thatU,d is a Branciari b-metric space with constant s=12/5>1.LetT:U⟶U be a mapping satisfying
(22)Tx=14,x∈A,15,x∈B.Now, we verify thatT is an F-contraction satisfying the condition F. We take x∈A, y∈B, and τ=0.1. Then, dTx,Ty=d1/4,1/5=1/400>0 and
(23)0.1+125⋅dTx,Ty=0.106<0.5=dx,y.LetF:0,∞⟶ℝ be a mapping defined by Fx=x, then it is easy to see that F∈Fs,τ. In addition, we can get
(24)τ+Fs⋅dTx,Ty≤Fdx,y.Hence,T satisfies the conditions of Theorem 8 and x=1/4 is the unique fixed point of T.


## 3.2. Weak Well-Posedness for the Fixed-Point Problem of anF-Contraction

The notion of well-posedness of a fixed-point problem has evoked much interest to several researchers [21–23]. Now, we introduce the concept of weak well-posedness for the fixed-point problem of the mapping T.Definition 9.
LetU,d be a metric space and T:U⟶U be a mapping. The fixed-point problem of T is said to be weakly well-posed if
(1)
T has a unique fixed point x∗ in U(2)
For any sequencexnn∈ℕ in U with limn→∞dxn,Txn=0 andlimn,m→∞dTxn,Txm=0, we have limn→∞dxn,x∗=0Theorem 10.
LetU,d be a complete Branciari b-metric space with constant s≥1 and T:U⟶U be an F-contraction for F∈Fs,τ with the continuous differentiable function F′ and limx→∞F′x<∞. Then, the fixed-point problem of T is weakly well-posed.Proof.
From Theorem8, it follows that T satisfies (5). Let x∗ be the unique fixed point of T. For any sequence xnn∈ℕ in U with limn→∞dxn,Txn=0 and limn,m→∞dTxn,Txm=0, in general, the sequence xnn∈ℕ can be decomposed into two subsequences xn=ynkk∈ℕ∪znkk∈ℕ, where Tynk=x∗ and Tznk≠x∗ for any k∈ℕ. Consequently, we can prove the following two assertions: Fdxn,x∗−FsdTxn,Tx∗≥τ.
(W1) ifxnn∈ℕ is such that Txn=x∗ for all n∈ℕ, then limn→∞dxn,x∗=0
(W2) ifxnn∈ℕ is such that Txn≠x∗ for all n∈ℕ, then limn→∞dxn,x∗=0In fact, (W1) is trivial, sincedxn,x∗=dxn,Txn⟶0 as n⟶∞. Now, we prove (W2). Because of the uniqueness of the fixed point of T, we can assume that there exists n0∈ℕ such that Txn≠xn for n≥n0. If Txn≠x∗ for any n∈ℕ, then we can apply F to get
(25)Fdxn,x∗−FsdTxn,Tx∗≥τ.SinceF is strictly increasing, we have dxn,x∗>sdTxn,Tx∗. Then,
(26)Fdxn,x∗−FsdTxn,Tx∗dxn,x∗−sdTxn,Tx∗≥τdxn,x∗−sdTxn,Tx∗.Then there exists a constantcnbetween dxn,x∗ and sdTxn,Tx∗ such that
(27)F′cn=Fdxn,x∗−FsdTxn,Tx∗dxn,x∗−sdTxn,Tx∗≥τdxn,x∗−sdTxn,Tx∗,which implies
(28)0<τF′cn≤dxn,x∗−sdTxn,Tx∗.Actually, we can assume that there exists a distinct subsequenceTxnk of {Txn}. Otherwise, there exists x0∈U and n1∈ℕ such that Txn=x0 for n≥n1. Since limn→∞dxn,Txn=0, we get limn→∞dxn,x0=0. If x0≠x∗, then x0≠Tx0 because of the uniqueness of the fixed point of T. For n≥n1, we obtain x0=Txn≠Tx0. Since T is an F-contraction, together with (19), we have
(29)0<dTxn,Tx0≤dxn,x0⟶0asn⟶∞,which is a contradiction. Hence, there exist m,q,n>n0 (m>q>n) such that Txm≠Txq≠Txn≠xn. Then,
(30)dxn,x∗≤sdxn,Txm+dTxm,Txn+dTxn,x∗.Combining with (28), we deduce that
(31)0<τF′cn≤sdxn,Txm+dTxm,Txn.Since(32)0≤dxn,Txm≤sdxn,Txn+dTxn,Txq+dTxq,Txm⟶0asn⟶∞.Then,(33)0<τF′cn≤sdxn,Txm+dTxm,Txn⟶0asn⟶∞.Therefore, we getlimn→∞F′cn=∞ which implies limn→∞cn=0 since F′ is continuous and limx→∞F′x<∞. Hence, we deduce that limn→∞sdTxn,Tx∗=0. From (30), it follows that limn→∞dxn,x∗=0.


## 3.3. The Problem of Common Fixed Points for TwoF-Contractions

Now, we continue to solve the problem of common fixed points for twoF-contraction mappings in a complete Branciari b-metric space.Theorem 11.
LetU,d be a complete Branciari b-metric space with constant s>1. If there exist τ>0 and F∈Fs,τ, such that S,T:U⟶U are two F-contraction mappings on U and satisfy
(34)τ+FdSx,Ty≤Fadx,y+bdx,Sx+cdy,Ty,for any a,b,c∈0,1 with a+b+c<1, sb<1, and mindSx,Ty,dx,y,dx,Sx,dy,Ty>0 for any x,y∈U×U. Then, there exists a unique common fixed point of the pair mappings S,T.Proof.
Letx0∈U, we define xn⊆U by
(35)Sx2j=x2j+1,Tx2j+1=x2j+2,j=0,1,2,⋯.
Combining with (34) and (35), we have
(36)τ+Fdx2j+1,x2j+2=τ+FdSx2j,Tx2j+1≤Fadx2j,x2j+1+bdx2j,Sx2j+cdx2j+1,Tx2j+1=Fadx2j,x2j+1+bdx2j,x2j+1+cdx2j+1,x2j+2.
Letλ=a+b/1−c, it is not difficult to see 0<λ<1 since a+b+c<1. Thanks to the strictly monotone increasing property of F, we deduce
(37)dx2j+1,x2j+2<a+b1−cdx2j,x2j+1=λdx2j,x2j+1.
Similarly,(38)dx2j+2,x2j+3<a+b1−cdx2j+1,x2j+2=λdx2j+1,x2j+2.
Hence,(39)dxn,xn+1<λdxn−1,xn,n∈ℕ.
For anyn∈ℕ, we obtain
(40)dxn,xn+1<λdxn−1,xn<λ2dxn−2,xn−1<⋯<λndx0,x1.
Notice that(41)τ+Fsdx1,x3=τ+FsdSx0,Sx2≤Fdx0,x2,τ+Fsdx2,x4=τ+FsdTx1,Tx3≤Fdx1,x3.
SinceF is strictly monotone increasing, we have
(42)dx1,x3≤1sdx0,x2,dx2,x4≤1sdx1,x3≤1s2dx0,x2.
By induction, we get(43)dxn,xn+2<1sndx0,x2,n∈ℕ.Next, we will show thatxn is a Cauchy sequence in U. We consider the following two cases.
(i)
Letm=n+p, if p is odd and p>2, we have(44)dxn,xm≤sdxn,xn+1+dxn+1,xn+2+dxn+2,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s2dxn+4,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s3dxn+4,xn+5+s3dxn+5,xn+6+⋯+sm−n/2dxm−2,xm−1+sm−n/2dxm−1,xm≤sλndx0,x1+sλn+1dx0,x1+s2λn+2dx0,x1+s2λn+3dx0,x1+s3λn+3dx0,x1+s3λn+4dx0,x1+⋯+sm−n/2λm−2dx0,x1+sm−n/2λm−1dx0,x1≤sλn+s2λn+2+s3λn+4+⋯+sm−n/2λm−2dx0,x1+sλn+1+s2λn+3+s3λn+5+⋯+sm−n/2λm−1dx0,x1≤sλn+s2λn+2+s3λn+4+⋯+sm−n/2λm−21+λdx0,x1≤sλn1+sλ2+s2λ4+s3λ6+⋯+sm−n−2/2λm−n−21+λdx0,x1≤sλn⋅1−sm−n/2λm−n1−sλ21+λdx0,x1≤sλn⋅1−sp/2λp1−sλ21+λdx0,x1.(ii)
Letm=n+p, if p is even and p>2, we have(45)dxn,xm≤sdxn,xn+1+dxn+1,xn+2+dxn+2,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s2dxn+4,xm≤sdxn,xn+1+sdxn+1,xn+2+s2dxn+2,xn+3+s2dxn+3,xn+4+s3dxn+4,xn+5+s3dxn+5,xn+6+⋯+sm−n−2/2dxm−4,xm−3+sm−n−2/2dxm−3,xm−2+sm−n−2/2dxm−2,xm≤sλndx0,x1+sλn+1dx0,x1+s2λn+2dx0,x1+s2λn+3dx0,x1+s3λn+3dx0,x1+s3λn+4dx0,x1+⋯+sm−n−2/2λm−4dx0,x1+sm−n−2/2λm−3dx0,x1+sm−n−2/2dxm−2,xm≤sλn+s2λn+2+s3λn+4+⋯+sm−n−2/2λm−4dx0,x1+sλn+1+s2λn+3+s3λn+5+⋯+sm−n−2/2λm−3dx0,x1+sm−n−2/2dxm−2,xm≤sλn+s2λn+2+s3λn+4+⋯+sm−n−2/2λm−41+λdx0,x1+sm−n−2/2dxm−2,xm≤sλn1+sλ2+s2λ4+s3λ6+⋯+sm−n−4/2λm−n−41+λdx0,x1+sm−n−2dxm−2,xm≤sλn⋅1−sm−n−2/2λm−n−21−sλ21+λdx0,x1+s−ndx0,x2≤sλn⋅1−sp−2/2λp−21−sλ21+λdx0,x1+s−ndx0,x2.Lettingm,n⟶∞, we obtain that dxn,xm⟶0 for all p>2. Hence, xn is a Cauchy sequence in U. From the completeness of U, it follows that there exists z∗∈U such that
(46)limn→∞dxn,z∗=0.Next, we verify thatz∗ is a fixed point of the mapping S. Indeed, if dSz∗,z∗>0, then
(47)τ+FdSz∗,x2j+2≤Fadz∗,x2j+1+bdz∗,Sz∗+cdx2j+1,x2j+2.By using the strictly monotone increasing property ofF, we get
(48)dSz∗,x2j+2<adz∗,x2j+1+bdz∗,Sz∗+cdx2j+1,x2j+2.We can also see that
(49)dSz∗,z∗<sdSz∗,x2j+2+dx2j+2,x2j+1+dx2j+1,z∗.It follows that(50)1sdSz∗,z∗≤liminfj→∞dSz∗,x2j+2≤limsupj→∞dSz∗,x2j+2≤bdz∗,Sz∗.Hence, we deduce that1/s≤b which contradicts the fact sb<1. Therefore, we obtain dSz∗,z∗=0, i.e., Sz∗=z∗. Similarly, we can get Tz∗=z∗. Therefore, we have
(51)Tz∗=Sz∗=z∗.Next, we will show thatS and T have a unique common fixed point. Suppose that z∗, z∗∗ are two distinct common fixed points of S and T. Then,
(52)τ+Fdz∗,z∗∗=τ+FdSz∗,Tz∗∗≤Fadz∗,z∗∗+bdz∗,Sz∗+cdz∗∗,Tz∗∗=Fadz∗,z∗∗+bdz∗,z∗+cdz∗∗,z∗∗.In view of the strictly monotone increasing property ofF, we obtain 1−adz∗,z∗∗<0, which is a contradiction. It means z∗=z∗∗.Next, we discuss the existence of fixed points for Hardy-Rogers-type set-valuedF-contractions in a complete Branciari b-metric space.Lemma 12 (see [24]).
IfF is right continuous and satisfies (F1), then
(53)FinfA=infFA,for all A∈0,∞ with infA>0.Definition 13 (see [24]).
LetU,d be a Branciari b-metric space and CBU be the family of all nonempty closed and bounded subsets of U. Let H:CBU×CBU⟶0,∞ be a function defined by HA,B=maxsupx∈Adx,B,supy∈Bdy,A for all A,B∈CBU, where dx,B=infdx,y:y∈B. Then, H defines a metric on CBU called the Hausdorff metric induced by d.Theorem 14.
LetU,d be a complete Branciari b-metric space with constant s≥1. Assume that there exist τ>0 and F∈Fs,τ such that T:U⟶CBU satisfies
(54)2τ+FHTx,Ty≤Fadx,y+bdx,Tx+cdy,Ty,for all x,y∈U and a,b,c≥0 with a+b+c<1 and sc<1. If F is right continuous, then T has a fixed point in U.Proof.
Letx0 be an arbitrary point of U and choose x1∈Tx0. If x1∈Tx1, then x1 is a fixed point of T and the proof is finished. Assume that x1∉Tx1, then Tx0≠Tx1. Since F is right continuous and Tx1 is closed, it follows from Lemma 12 that
(55)Fdx1,Tx1=infy∈Tx1Fdx1,y.
Noticing thatdx1,Tx1≤HTx0,Tx1, then there exists x2∈Tx1 such that
(56)Fdx1,x2−τ<Fdx1,Tx1≤FHTx0,Tx1.
Consequently, we get(57)Fdx1,x2<FHTx0,Tx1+τ,which implies
(58)2τ+Fdx1,x2<2τ+FHTx0,Tx1+τ≤Fadx0,x1+bdx0,Tx0+cdx1,Tx1+τ≤Fadx0,x1+bdx0,x1+cdx1,x2+τ=Fa+bdx0,x1+cdx1,x2+τ.
SinceF is strictly increasing, we deduce
(59)dx1,x2<a+bdx0,x1+cdx1,x2,and hence,
(60)1−cdx1,x2<a+bdx0,x1.
Sincea+b+c<1, we obtain
(61)dx1,x2<a+b1−cdx0,x1<dx0,x1.
Combining with (58), we get
(62)τ+Fdx1,x2<Fdx0,x1.
By induction, we can get a sequencexn⊂U satisfying dxn,xn+1 is decreasing, xn∉Txn, xn+1∈Txn, and τ+Fdxn,xn+1<Fdxn−1,xn for all n∈ℕ.
Letγn=dxn,xn+1 and βn=dxn,xn+2. Then, we have
(63)Fγn+1<Fγn−τ,for all n∈ℕ. Similarly to the proof of Theorem 8 (while s=1), we deduce that xn is a Cauchy sequence. Since U is complete, there exists x∗∈U such that xn⟶x∗ as n⟶∞. Now, we prove that x∗ is a fixed point of T. Since
(64)2τ+Fdxn,Tx∗≤2τ+FHTxn−1,Tx∗≤Fadxn−1,x∗+bdxn−1,Txn−1+cdx∗,Tx∗,dxn−1,Txn−1≤dxn−1,xn,we deduce that
(65)dxn,Tx∗<adxn−1,x∗+bdxn−1,xn+cdx∗,Tx∗.
It follows that(66)limsupn→∞dxn,Tx∗≤cdx∗,Tx∗.
Noting that(67)dx∗,Tx∗=infy∈Tx∗dx∗,y≤sinfy∈Tx∗dx∗,xn+1+dxn+1,xn+dxn,y≤sdx∗,xn+1+dxn+1,xn+sdxn,Tx∗,we obtain
(68)dx∗,Tx∗≤slimsupn→∞dxn,Tx∗.
Combining with (66) and (68), we get
(69)limsupn→∞dxn,Tx∗≤sclimsupn→∞dxn,Tx∗.
Sincesc<1, then limsupn→∞dxn,Tx∗=0 which implies dx∗,Tx∗=0. Hence, x∗ is a fixed point of T since Tx∗ is closed.


## 4. Applications

### 4.1. Existence of Solutions of Integral Equations

Firstly, we will apply Theorem8 to solving the existence and uniqueness of the solution of the following integral equation:
(70)xt=gt+∫0amt,ufu,xudu,t∈0,a,a>0,where m:0,a×0,a⟶0,∞ is a function such that mt,⋅∈L10,a, f:0,a×ℝ⟶ℝ, and g:0,a⟶ℝ are two continuous functions.LetU=C0,a,ℝ be the set of real continuous functions defined on 0,a and the operator T:U⟶U be defined by
(71)Txt=gt+∫0amt,ufu,xudu,t∈0,a.We will prove the following result.Theorem 15.
LetU=C0,a,ℝ and d:U×U⟶0,∞ be given by dx,y=supt∈0,axt−yt2 for all x,y∈U.Suppose there existsζ:U×U⟶0,3/3 such that
(72)fx,xu−fy,yu≤ζx,yxu−yuand ∫0amt,udu≤e−τ for u∈0,a and τ>0. Then, the integral equation (70) has a unique solution in X.Proof.
Clearly, each fixed point ofT in (71) is a solution of (70). Meanwhile, it is not difficult to see that U,d is a complete Branciari b-metric space with s=3. Then, we obtain
(73)Txt−Tyt2=∫0amt,ufu,xu−fu,yudu2≤∫0amt,uζx,yxu−yudu2≤13∫0amt,udu2supu∈0,axu−yu2≤e−τ3dx,y.
Consequently, for eachx,y∈U, we have
(74)3dTx,Ty≤e−τdx,y,which implies
(75)τ+ln3dTx,Ty≤lndx,y,for each x,y∈U. From Definition 7, the mapping T:U⟶U is an F-contraction with Fx=lnx. By applying Theorem 8, the operator T has a unique fixed point, that is, the integral equation (70) has a unique solution.


### 4.2. Existence of the Common Solution of Functional Equations

Finally, we will apply Theorem11 to solving the existence and uniqueness of the solution of the dynamic programming problem. In particular, the problem of dynamic programming related to multistage process reduces to solving the existence and uniqueness of the solution of the following functional equations:
(76)pu=supv∈Bfu,v+G1u,v,pφu,v,forallu∈A,(77)qu=supv∈Bfu,v+G2u,v,pφu,v,forallu∈A,where f:A×B⟶ℝ, φ:A×B⟶A, and G1,G2:A×B×ℝ⟶ℝ. We assume that C and D are Banach spaces, A⊆C is a state space, and B⊆D is a decision space. For details, see also [25].In this section, we will show that the equations (76) and (77) have one common solution. Let U be the set of all bounded real-valued functions defined on a nonempty set A. We define the norm ⋅ by
(78)x=supu∈Axu,for all x∈U, it is not difficult to see that U,⋅ is a Banach space. Moreover, we define
(79)dx,y=supu∈Axu−yu2,for all x,y∈U. Then, U,d is a complete Branciari b-metric space with s=3. Consider the operators S,T:U⟶ℝ given by
(80)Sxu=supv∈Bfu,v+G1u,v,xφu,v,forallu∈A,(81)Tyu=supv∈Bfu,v+G2u,v,yφu,v,forallu∈A,Now, we prove the following theorem.Theorem 16.
LetS,T:U⟶ℝ be defined as in (80) and (81), respectively. Suppose that the following hypotheses hold:
(i)
f:A×B⟶ℝ and G1,G2:A×B×ℝ⟶ℝ are bounded functions(ii)
There existsλ>0 such that, for all u∈A, v∈B, and t,r∈ℝ(82)Giu,v,t−Giu,v,r≤e−λ3t−r,i=1,2(iii)
There existsτ>0 such that, for all u∈A, v∈B, and x,y∈U(83)G1u,v,xu−G2u,v,yu2≤e−τMx,y,where
(84)Mx,y=adx,y+bdx,Sx+cdy,Ty,for all a,b,c∈0,1 with a+b+c<1, 3b<1, and mindSx,Ty,dx,y,dx,Sx,dy,Ty>0Then, there exists a unique common fixed point of the pair mappingsS,T.Proof.
Indeed, letx∈U, then x is bounded, there exists λ1>0 such that xu≤λ1 for all u∈A. By hypothesis (i), there exist λ2,λ3, and λ4 such that
(85)fu,v≤λ2,G1u,v,0≤λ3,G2u,v,0≤λ4,for all u∈A and v∈B. Furthermore, by (ii), for all u∈A and v∈B, we have
(86)fu,v+G1u,v,xφu,v≤fu,v+G1u,v,xφu,v≤fu,v+G1u,v,xφu,v−G1u,v,0+G1u,v,0≤λ2+e−λ3xφu,v+λ3≤λ2+e−λ3λ1+λ3.
Consequently, for allu∈A, v∈B, and x,y∈U, we get
(87)Sxu=supv∈Bfu,v+G1u,v,xφu,v≤λ2+e−λ3λ1+λ3.
Similarly,(88)Txu=supv∈Bfu,v+G2u,v,xφu,v≤λ2+e−λ3λ1+λ4,
which show thatS and T are bounded. Let ε>0, x1,x2∈U, u∈A, there exist v1,v2∈B such that
(89)Sx1u<fu,v1+G1u,v1,x1φu,v1+ε,(90)Tx2u<fu,v2+G2u,v2,x2φu,v2+ε,(91)Sx1u<fa,v2+G1u,v2,x1φu,v2,(92)Tx2u<fu,v1+G2u,v1,x2φu,v1.
Then by (89) and (92), we obtain
(93)Sx1u−Tx2u<G1u,v1,x1φu,v1−G2u,v1,x2φu,v1+ε,≤G1u,v1,x1φu,v1−G2u,v1,x2φu,v1+ε.
Similarly, by (90) and (91), we deduce
(94)Tx2u−Sx1u<G2u,v1,x2φu,v1−G1u,v1,x1φu,v1+ε.
Besides, we also find(95)Sx1u−Sx2u<G1u,v1,x1φu,v1−G1u,v1,x2φu,v1+ε≤e−λ3x1φu,v1−x2φu,v1+ε.
Sinceε is arbitrary, we conclude
(96)Sx1u−Tx2u2≤G1u,v1,x1φu,v1−G1u,v1,x2φu,v12≤e−τMx1,x2,Sx1u−Sx2u2≤e−λ3x1φu,v1−x2φu,v12.
Hence, we deduce(97)dSx1,Tx2≤e−τMx1,x2,3dSx1,Sx2≤e−λdx1,x2.
Similarly, we can get(98)3dTx1,Tx2≤e−λdx1,x2.
Then,(99)τ+lndSx1,Tx2≤lnMx1,x2,λ+ln3dSx1,Sx2≤lndx1,x2,λ+ln3dTx1,Tx2≤lndx1,x2,
which show thatS and T are two F-contraction mappings on U with Fx=lnx. By Theorem 11, there exists a unique common fixed point of the pair mappings S,T which is the solution of the equations (76) and (77).


## 4.1. Existence of Solutions of Integral Equations

Firstly, we will apply Theorem8 to solving the existence and uniqueness of the solution of the following integral equation:
(70)xt=gt+∫0amt,ufu,xudu,t∈0,a,a>0,where m:0,a×0,a⟶0,∞ is a function such that mt,⋅∈L10,a, f:0,a×ℝ⟶ℝ, and g:0,a⟶ℝ are two continuous functions.LetU=C0,a,ℝ be the set of real continuous functions defined on 0,a and the operator T:U⟶U be defined by
(71)Txt=gt+∫0amt,ufu,xudu,t∈0,a.We will prove the following result.Theorem 15.
LetU=C0,a,ℝ and d:U×U⟶0,∞ be given by dx,y=supt∈0,axt−yt2 for all x,y∈U.Suppose there existsζ:U×U⟶0,3/3 such that
(72)fx,xu−fy,yu≤ζx,yxu−yuand ∫0amt,udu≤e−τ for u∈0,a and τ>0. Then, the integral equation (70) has a unique solution in X.Proof.
Clearly, each fixed point ofT in (71) is a solution of (70). Meanwhile, it is not difficult to see that U,d is a complete Branciari b-metric space with s=3. Then, we obtain
(73)Txt−Tyt2=∫0amt,ufu,xu−fu,yudu2≤∫0amt,uζx,yxu−yudu2≤13∫0amt,udu2supu∈0,axu−yu2≤e−τ3dx,y.
Consequently, for eachx,y∈U, we have
(74)3dTx,Ty≤e−τdx,y,which implies
(75)τ+ln3dTx,Ty≤lndx,y,for each x,y∈U. From Definition 7, the mapping T:U⟶U is an F-contraction with Fx=lnx. By applying Theorem 8, the operator T has a unique fixed point, that is, the integral equation (70) has a unique solution.


## 4.2. Existence of the Common Solution of Functional Equations

Finally, we will apply Theorem11 to solving the existence and uniqueness of the solution of the dynamic programming problem. In particular, the problem of dynamic programming related to multistage process reduces to solving the existence and uniqueness of the solution of the following functional equations:
(76)pu=supv∈Bfu,v+G1u,v,pφu,v,forallu∈A,(77)qu=supv∈Bfu,v+G2u,v,pφu,v,forallu∈A,where f:A×B⟶ℝ, φ:A×B⟶A, and G1,G2:A×B×ℝ⟶ℝ. We assume that C and D are Banach spaces, A⊆C is a state space, and B⊆D is a decision space. For details, see also [25].In this section, we will show that the equations (76) and (77) have one common solution. Let U be the set of all bounded real-valued functions defined on a nonempty set A. We define the norm ⋅ by
(78)x=supu∈Axu,for all x∈U, it is not difficult to see that U,⋅ is a Banach space. Moreover, we define
(79)dx,y=supu∈Axu−yu2,for all x,y∈U. Then, U,d is a complete Branciari b-metric space with s=3. Consider the operators S,T:U⟶ℝ given by
(80)Sxu=supv∈Bfu,v+G1u,v,xφu,v,forallu∈A,(81)Tyu=supv∈Bfu,v+G2u,v,yφu,v,forallu∈A,Now, we prove the following theorem.Theorem 16.
LetS,T:U⟶ℝ be defined as in (80) and (81), respectively. Suppose that the following hypotheses hold:
(i)
f:A×B⟶ℝ and G1,G2:A×B×ℝ⟶ℝ are bounded functions(ii)
There existsλ>0 such that, for all u∈A, v∈B, and t,r∈ℝ(82)Giu,v,t−Giu,v,r≤e−λ3t−r,i=1,2(iii)
There existsτ>0 such that, for all u∈A, v∈B, and x,y∈U(83)G1u,v,xu−G2u,v,yu2≤e−τMx,y,where
(84)Mx,y=adx,y+bdx,Sx+cdy,Ty,for all a,b,c∈0,1 with a+b+c<1, 3b<1, and mindSx,Ty,dx,y,dx,Sx,dy,Ty>0Then, there exists a unique common fixed point of the pair mappingsS,T.Proof.
Indeed, letx∈U, then x is bounded, there exists λ1>0 such that xu≤λ1 for all u∈A. By hypothesis (i), there exist λ2,λ3, and λ4 such that
(85)fu,v≤λ2,G1u,v,0≤λ3,G2u,v,0≤λ4,for all u∈A and v∈B. Furthermore, by (ii), for all u∈A and v∈B, we have
(86)fu,v+G1u,v,xφu,v≤fu,v+G1u,v,xφu,v≤fu,v+G1u,v,xφu,v−G1u,v,0+G1u,v,0≤λ2+e−λ3xφu,v+λ3≤λ2+e−λ3λ1+λ3.
Consequently, for allu∈A, v∈B, and x,y∈U, we get
(87)Sxu=supv∈Bfu,v+G1u,v,xφu,v≤λ2+e−λ3λ1+λ3.
Similarly,(88)Txu=supv∈Bfu,v+G2u,v,xφu,v≤λ2+e−λ3λ1+λ4,
which show thatS and T are bounded. Let ε>0, x1,x2∈U, u∈A, there exist v1,v2∈B such that
(89)Sx1u<fu,v1+G1u,v1,x1φu,v1+ε,(90)Tx2u<fu,v2+G2u,v2,x2φu,v2+ε,(91)Sx1u<fa,v2+G1u,v2,x1φu,v2,(92)Tx2u<fu,v1+G2u,v1,x2φu,v1.
Then by (89) and (92), we obtain
(93)Sx1u−Tx2u<G1u,v1,x1φu,v1−G2u,v1,x2φu,v1+ε,≤G1u,v1,x1φu,v1−G2u,v1,x2φu,v1+ε.
Similarly, by (90) and (91), we deduce
(94)Tx2u−Sx1u<G2u,v1,x2φu,v1−G1u,v1,x1φu,v1+ε.
Besides, we also find(95)Sx1u−Sx2u<G1u,v1,x1φu,v1−G1u,v1,x2φu,v1+ε≤e−λ3x1φu,v1−x2φu,v1+ε.
Sinceε is arbitrary, we conclude
(96)Sx1u−Tx2u2≤G1u,v1,x1φu,v1−G1u,v1,x2φu,v12≤e−τMx1,x2,Sx1u−Sx2u2≤e−λ3x1φu,v1−x2φu,v12.
Hence, we deduce(97)dSx1,Tx2≤e−τMx1,x2,3dSx1,Sx2≤e−λdx1,x2.
Similarly, we can get(98)3dTx1,Tx2≤e−λdx1,x2.
Then,(99)τ+lndSx1,Tx2≤lnMx1,x2,λ+ln3dSx1,Sx2≤lndx1,x2,λ+ln3dTx1,Tx2≤lndx1,x2,
which show thatS and T are two F-contraction mappings on U with Fx=lnx. By Theorem 11, there exists a unique common fixed point of the pair mappings S,T which is the solution of the equations (76) and (77).



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