# Fixed Point Results on Closed Ball in Convex Rectangularb−Metric Spaces and Applications

Authors: Chaobo Li; Yunan Cui; Lili Chen
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, the concept of convex rectangularb−metric spaces is introduced as a generalization of both convex metric spaces and rectangular b−metric spaces. The purpose of this study is to indicate a way of generalizing Mann’s iteration algorithm and a series of fixed point results in rectangular b−metric spaces. Furthermore, certain examples are given to support the results. We also study well posedness of fixed point problems of some mappings in convex rectangular b−metric spaces, and an application to the dynamic programming is entrusted to manifest the viability of the obtained results. Our results extend comparable results in the existing literature.


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## Body

## 1. Introduction and Preliminaries

It is well known that fixed point theory has become an important field of mathematics due to its high degree of unity and wide application. No doubt that the most significant fundamental result of this theory is Banach contraction principle [1] which was published in 1922. Banach contraction principle proposes for the first time to use Picard iteration to approximate a fixed point, which not only proves the existence of the fixed point but also proves the uniqueness of the fixed point. Later in 1968, Kannan [2] studied a new type of contractive mapping. Since then, there have been many results related to mappings satisfy various types of contractive inequality, see for example [3–9].In 2000, Branciari [10] developed the notion of a rectangular space as a generalization of normal metric space via substituting the triangle inequality with the quadrilateral inequality and extended Banach contraction principle to this space. Successively, George et al. [11] introduced the notion of a rectangular b−metric space as a generalization of rectangular metric space and they also proved some fixed point results for contractive mappings. The concept of a convex structure and a convex metric space was introduced by Takahashi [12]. Later, Goebel and Kirk [13] studied some iterative processes for nonexpansive mappings in a hyperbolic metric space, and in 1988, Ding [14] found fixed points of quasicontraction mappings in convex metric spaces by Ishikawa’s iteration scheme. However, iterative methods have received vast investigation for finding fixed points of nonexpansive mappings, see [15–17]. Particularly, in the process of the research on some fixed point problem, one of the most famous fixed point method is the Mann iteration [18, 19] as follows:
(1)xn+1=αnxn+1−αnTxn,for some suitably chosen scalars αn∈0,1. Due to [20], Mann iterative sequence xn converges weakly to a fixed point of T if the sequence αn∈0,1 satisfies following conditions: ∑n=1∞αn1−αn=∞.Very recently, Chen et al. [21] introduced the notion of a convex b−metric space and extend Mann’s algorithms directly to b−metric spaces. After that, Asif et al. [22] investigate fixed point of single-valued Hardy-Roger’s type F−contraction globally as well as locally in a convex b−metric space. Along the line, Chen et al. [23] introduce the concept of a convex graphical rectangular b−metric space and prove some fixed point theorems in this space. New some fixed point results on a closed ball can be seen in [22, 24–26].In this work, we firstly introduce the concept of the convex rectangularb−metric spaces which is a combination of properties of rectangular b−metric spaces and convex metric spaces. However, we prove some fixed point theorems using generalized Mann’s iteration algorithm and show concrete examples supporting our main results. In addition, we claim that fixed point problem is well posed and as an application, we apply our main results to solve the dynamic programming problem.Some fundamental definitions related to our work are given below:Definition 1. (See [11]).
LetX be a nonempty set and the mapping d:X×X⟶0,∞ satisfy
dx,y=0 if and only if x=y for all x,y∈Xdx,y=dy,x for all x,y∈X(3)
There exists a real number such thatdx,y≤sdx,u+du,v+dv,y for all x,y∈X and all distinct points u,v∈X\x,y
Then,d is called a rectangular b−metric on X, and X,d is called a rectangular b−metric space RbMS with coefficient s≥1.Remark 2.
Note that every metric space is a rectangular metric spaceRMS (see [11]), and every RMS is a RbMS with coefficient s=1.Definition 3. (See [11]).
LetX,d be a RbMS, xn be a sequence in X and x∈X. Then,
(a)
The sequencexn is said to be convergent in X to x, if for every ε>0, there exists n0∈ℕ such that dxn,x<ε for all n>n0, and this fact is represented by limn⟶∞xn=x(b)
The sequencexn is said to be a Cauchy sequence in X if for every ε>0, there exists n0∈ℕ such that dxn,xm<ε for all n,m>n0, and this fact is represented by limn,m⟶∞dxn,xm=0X is said to be a complete RbMS if every Cauchy sequence in X converges to some x∈XDefinition 4. (See [12]).
LetX,d be a metric space and I=0,1. A continuous function w:X×X×0,1⟶X is said to be a convex structure on X if for each x,y∈X and α∈I,
(2)du,wx,y,α≤αdu,x+1−αdu,y,for all u∈X. A metric space X,d with a convex structure w is called a convex metric space.


## 2. Main Results

In this section, we introduce a generalization of both convex metric spaces and rectangularb−metric spaces, which we call convex rectangular b−metric spaces. We also establish some fixed point theorems arising from this metric space.Definition 5.
LetX,d be a RbMS with constant s≥1. If a mapping w:X×X×0,1⟶X satisfies
(3)du,wx,y,α≤αdu,x+1−αdu,y,for all x,y,u∈X and α∈0,1, then X,d,w is said to be a convex rectangular b−metric space CRbMS.Definition 6.
LetX,d,w be a CRbMS and T:X⟶X be a mapping. Let xn be the sequence generated by Mann’s iterative procedure involving the mapping T, as follows:
(4)xn+1=wxn,Txn;αn,where αn∈0,1 and x0∈X are the initial value.Definition 7.
Ifs=1 in Definition 5, we call the resultant space to be a convex rectangular metric space CRMS, which is, indeed, the RMS with a convex structure w.Next, we see some specific examples ofCRbMS.Example 8.
LetX=ℝ. For any x,y∈X, we define the metric d:X×X⟶0,+∞ by dx,y=x−yr and r≥1. Notice that, for any a,b,c∈0,+∞ and 1≤r<∞, then the convex of the function fx=xrx>0 implies that
(5)a+b+c3r≤ar+br+cr3a,b,c>0.
Then, for any distinct pointsu,v∈X\x,y, we have
(6)dx,y=x−yr=x−u+u−v+v−yr≤x−u+u−v+v−yr≤3r−1x−ur+u−vr+v−yr=3r−1dx,u+du,v+dv,y.
Hence,X,d is a RbMS with s=3r−1. For any x,y∈X, let w:X×X×1/2⟶X be a mapping defined by
(7)wx,y;α=αx+1−αy,α=12.
Now, we verify thatw satisfies inequality (3). In fact, for any x,y,u∈X, we can see that
(8)du,wx,y;α=u−αx+1−αyr≤2r−1αru−xr+1−αru−yr=αdu,x+1−αdu,y.
Therefore,X,d,w is a CRbMS with s=3r−1. Note that X,d,w is not a metric space as follows:
(9)d1,3=22>d1,2+d2,3=2,for we take r=2. Moreover, X,d,w is a CRMS when we let r=1, and it shows that CRbMS reduces to a CRMS for s=1.Example 9.
LetX=ℝ. For any x,y∈X, we define the metric d:X×X⟶0,+∞ by dx,y=x−y2. From Example 8, it follows that X,d is a RbMS with s=3. For any x,y∈X, let w:X×X×0,1⟶X be a mapping defined by
(10)wx,y;α=αx+1−αy.
For anyx,y,u∈X, we obtain that
(11)du,wx,y;α=u−αx+1−αy2≤αu−x+1−αu−y2=α2du,x+1−α2du,y+2α1−αu−xu−y≤α2du,x+1−α2du,y+2α1−αu−x2+u−y22=α2du,x+1−α2du,y+α1−αdu,x+du,y=αdu,x+1−αdu,y.
Therefore,X,d,w is a CRbMS with s=3, but not a CRMS.Example 10.
LetX=0,2, d:X×X⟶0,+∞, such that dx,y=dy,x and
(12)dx,y=0ifx=y,2aifx,y∈0,1,12aotherwise,where a>0 is a constant. Then, X,d is a RbMS with coefficient s=4/3. The mapping w:X×X×0,1⟶X is defined by wx,y;α=2−αxy, α=1/4, and then
(13)du,wx,y;α≤14du,x+34du,y.So,X,d,w is a CRbMS with coefficient s=4/3, but not a CRMS.Definition 11.
LetX,d,w be a CRbMS with constant s≥1, x0 is some element in X, and ε>0, and then the set Bεx0¯=x∈X:dx0,x≤ε is called a closed ball in X.In the paper [3], George et al. proved Banach contraction principle in complete RbMS by means of Picard iteration. Now, we will show Banach contraction principle for complete CRbMS using generalized Mann’s iteration algorithm.Theorem 12.
LetX,d,w be a complete CRbMS with constant s≥1 and T:X⟶X be a mapping satisfying
(14)dTx,Ty≤βdx,y,for all x,y∈X, where β∈0,1. Let the sequence xn generated by the Mann iterative process and x0∈X such that dx0,Tx0=M<∞. If β∈0,1/2s2 and αn∈0,sr/2s3+s−2 (r is an arbitrary positive real number and r<1), then T has a unique fixed point in X. Moreover, the sequence xn⊆Bεx0¯ and xn⟶x∗∈Bεx0¯ as n⟶∞, if the following inequality holds:
(15)dx0,Tx0≤β1−sβε.Proof.
Without loss of generality, we suppose thatxn≠xn+1 for all n∈ℕ. Indeed, If xn=xn+1, then xn=wxn,Txn;αn. We conclude that
(16)dxn,Txn=dwxn,Txn;αn,Txn≤αndxn,Txn,and it shows dxn,Txn=0; then, xn is a fixed point of T, and the proof is finished. It follows from Definition 5 and Definition 6,
(17)dxn,xn+1=dxn,wxn,Txn;αn≤1−αndxn,Txn.Now, we consider the following two cases:Case 13.
Ifxn≠Txn−1 for all n∈ℕ, we have
(18)dxn,Txn=dwxn−1,Txn−1;αn−1,Txn≤αn−1dxn−1,Txn+1−αn−1dTxn−1,Txn≤αn−1dxn−1,Txn+1−αn−1βdxn−1,xn≤sαn−1dxn−1,xn+dxn,Txn−1+dTxn−1,Txn+1−αn−12βdxn−1,Txn−1≤sαn−11−αn−1dxn−1,Txn−1+αn−1dxn−1,Txn−1+1−αn−1βdxn−1,Txn−1+1−αn−12βdxn−1,Txn−1≤sαn−11+1−αn−1β+1−αn−12βdxn−1,Txn−1.
Letλn−1=sαn−11+1−αn−1β+1−αn−12β, with the assumption 0≤β≤1/2s2 and 0≤αn≤sr/2s3+s−2, and we obtain that
(19)λn−1≤sαn−11+1−αn−112s2+1−αn−1212s2=sαn−1+αn−12s−αn−122s+12s2−αn−1s2+αn−122s2=12s2−12sαn−12+s+12s−1s2αn−1+12s2≤s+12s−1s2αn−1+12s2=2s3+s−22s2×sr2s3+s−2+12s≤r+12s≤r+12.
Hence,(20)dxn,Txn≤λn−1dxn−1,Txn−1≤r+12dxn−1,Txn−1.Case 14.
Ifxn=Txn−1 for some n∈ℕ, we have
(21)dxn,Txn=dTxn−1,Txn≤βdxn−1,xn≤β1−αndxn−1,Txn−1≤12s2dxn−1,Txn−1.
Denote thatλ=r+1/2<1, and it follows from (20) and (21) that
(22)dxn,Txn≤λdxn−1,Txn−1,foralln∈ℕ,which implies that dxn,Txn is a decreasing sequence of nonnegative reals. Hence, there exists γ≥0 such that
(23)limn⟶∞dxn,Txn=γ.We will show thatγ=0. Suppose that γ>0, letting n⟶∞ in inequality (22), we obtain
(24)γ≤λγ,a contradiction. Hence, we get that γ=0. Furthermore, we have
(25)dxn,xn+1≤1−αndxn,Txn,which shows that limn⟶∞dxn,xn+1=0. Also, we can assume xn≠xn+p for any p>1. Indeed, if xn=xn+p, then using the inequality (21), we have
(26)dxn,Txn=dxn+p,Txn+p≤λp−1dxn,Txn,in which shows that dxn,Txn=0 and xn=Txn, and then xn is a fixed point, and the proof is finished. Next, we shall prove limn⟶∞dxn,xn+2=0 for all n∈ℕ. In order to do it, we will consider the following two cases:Case 15.
Ifxn+2≠Txn for all n∈ℕ, then we have
(27)dxn,xn+2≤sdxn,Txn+dTxn,Txn+2+dTxn+2,xn+2≤sdxn,Txn+βdxn,xn+2+dTxn+2,xn+2,which establishes that
(28)dxn,xn+2≤s1−sβdxn,Txn+dxn+2,Txn+2as0≤β<12s2.Case 16.
If there exist somen∈ℕ such that xn+2=Txn, then
(29)dxn,xn+2≤dxn,Txn.
It follows from (28) and (29) that
(30)limn⟶∞dxn,xn+2=0.
Next, we claim thatxn is a Cauchy sequence by contradiction. Assume that there exists ε0>0 and the subsequences xθk and xηk of xn such for θk>ηk>k with dxθk,xηk≥ε0, dxθk−1,xηk<ε0. On the one hand,
(31)ε0≤dxθk,xηk≤sdxθk,xηk+1+dxηk+1,xηk+2+dxηk+2,xηk,taking the limit superior in above inequality as k⟶∞, and we conclude
(32)ε0s≤limsupk⟶∞dxθk,xηk+1.On the other hand, letxηk≠xηk+2≠xθk−1≠xηk+1 and Txηk≠xηk≠Txθk−1≠xηk+1, and we have
(33)dxθk,xηk+1=dwxθk−1,Txθk−1;αθk−1,xηk+1≤αθk−1dxθk−1,xηk+1+1−αθk−1dTxθk−1,xηk+1≤αθk−1sdxθk−1,xηk+dxηk,xηk+2+dxηk+2,xηk+1+1−αθk−1sdTxθk−1,Txηk+dTxηk,xηk+dxηk,xηk+1≤αθk−1s+1−αθk−1sβdxθk−1,xηk+αθk−1sdxηk,xηk+2+dxηk+2,xηk+1+1−αθk−1sdTxηk,xηk+dxηk,xηk+1,by taking the limit superior on both sides of above the inequality as k⟶∞, and we get
(34)ε0s≤limsupk⟶∞dxθk,xηk+1≤αθk−1s+1−αθk−1sβε0≤αθk−1s+12s1−αθk−1ε0≤αθk−1s+12s−αθk−12sε0≤12s2−1×s−12s+12sε0=r+12ε0s<1sε0,a contradiction. Thus, xn is a Cauchy sequence in X. Since the space X,d,w is complete, there exists x∗∈X such that limn⟶∞xn=x∗. We shall show that x∗ is a fixed point of T. Applying the rectangular inequality, we obtain that
(35)dx∗,Tx∗≤sdx∗,xn+dxn,xn+1+dxn+1,Tx∗≤sdx∗,xn+sdxn,xn+1+sαndxn,Tx∗+1−αndTxn,Tx∗≤sdx∗,xn+sdxn,xn+1+s2αndxn,xn+1+dxn+1,x∗+dx∗,Tx∗+s1−αnβdxn,x∗,since s2αn<1, and then
(36)dx∗,Tx∗≤11−s2αnsdx∗,xn+sdxn,xn+1+s2αndxn,xn+1+dxn+1,x∗+s1−αnβdxn,x∗,letting n⟶∞, and we deduce dx∗,Tx∗=0 which implies Tx∗=x∗. Thus, x∗ is a fixed point of T. Suppose that x∗,y∗∈X are two distinct fixed points of T, that is, Tx∗=x∗ and Ty∗=y∗. Then,
(37)dx∗,y∗=dTx∗,Ty∗≤βdx∗,y∗,which is a contradiction. Therefore, we must have dx∗,y∗=0, i.e., x∗=y∗. Thus, T has a unique fixed point. Next, we proceed to show that the sequence xn⊆Bεx0¯. In order to complete it, we will use mathematical induction. Thanks to Definition 5 and Definition 6, we obtain
(38)dx0,x1=dx0,wx0,Tx0;α0≤1−α0dx0,Tx0≤1−α0β1−sβε<ε,which implies dx0,x1<ε; therefore, x1∈Bεx0¯. Suppose x2,x3,⋯,xm∈Bεx0¯, observe from above proof, we get dxn,Txn≤λndx0,Tx0 for all n∈ℕ. It is easy to see that β1−sβ≤1/4s. Now, we can assume that xm+1≠xm. If Tx0≠Txm≠x0≠xm+1, then
(39)dx0,xm+1≤sdx0,Tx0+dTx0,Txm+dTxm,xm+1≤sβ1−sβε+βdx0,xm+αmdTxm,xm≤sβ1−sβε+βε+αmλmβ1−sβε<ε4+ε2+ε4=ε.We also need to distinguish the following four cases:Case 17.
Ifx0=Tx0, then we have
(40)dx0,xm+1=dx0,wxm,Txm;αm≤αmdx0,xm+1−αmdTx0,Txm≤αmε+1−αmβε<ε.Case 18.
ifx0=Txm, then we have
(41)dx0,xm+1=dTxm,wxm,Txm;αm≤αmdTxm,xm≤αmλmdTx0,x0≤αmλmβ1−sβε<ε.Case 19.
ifxm+1=Tx0, then we have
(42)dx0,xm+1=dx0,Tx0≤β1−βε<ε.Case 20.
ifxm+1=Txm, then we have
(43)dx0,xm+1=dx0,wxm,Txm;αm≤αmdx0,xm+1−αmdx0,Txm=αmdx0,xm+1−αmdx0,xm+1,which implies
(44)dx0,xm+1≤dx0,xm≤ε.Finally, by above cases, we prove thatdx0,xm+1≤ε which show that xm+1∈Bεx0¯. Hence, by induction xn∈Bεx0¯, therefore, we conclude that xn∈Bεx0¯ for all n∈ℕ. As every closed ball in a complete metric space is complete, so xn⟶x∗∈Bεx0, as n⟶∞.The following example illustrates the above theorem.Example 21.
LetX=R+∪0 and Tx=x/5 for all x∈X. For any x,y∈X, we define d:X×X⟶0,+∞ by dx,y=x−y2. The mapping w:X×X×0,1⟶X is defined by
(45)wx,y;α=αx+1−αy,x,y∈X.
Setxn+1=wxn,Txn;αn and αn=1/2s2+2. If β=1/2s2+1, then xn∈Bεx0¯ and T have a unique fixed point in Bεx0¯.Proof.
It is easy to see thatX,d is a CRbMS with s=3. In addition, for any x,y,u∈X, we have
(46)du,wx,y;αn=αnu−x+1−αnu−y2≤αn2u−x2+1−αn2u−y2+2αn1−αnu−xu−y≤αn2u−x2+1−αn2u−y2+αn1−αnu−x2+u−y2≤αnu−x2+1−αnu−y2.
So,X,d,w is a CRbMS with s=3. It is not difficult to see that T satisfies
(47)dTx,Ty=125dx,y≤βdx,y,for β=1/18. According to xn+1=wxn,Txn;αn, we have xn+1=1/20xn+19/20Txn, since Tx=x/5, and we obtain
(48)xn+1=120xn+1920×15xn,that is, xn+1=6/25xn, then
(49)xn=625xn−1,xn−1=625xn−2,⋯,x1=625x0,And we obtain
(50)xn=625nx0,Txn=15×625nx0,while n⟶∞, getting xn⟶0∈X and Txn⟶0∈X. Hence, 0 is a fixed point of T in X. Suppose x∗,y∗∈X are two distinct fixed points of T, then we have dx∗,y∗=dTx∗,Ty∗=1/25dx∗,y∗ which shows that dx∗,y∗=0, that is, x∗=y∗. Thus, T has a unique fixed point in X. Let ε=x02/β1−sβ>0, then β1−sβε=x02≥dx0,Tx0=16/25x02. For all n∈ℕ, we obtain dx0,xn=x0−6/25nx02<x02<ε, and this means that the sequence xn⊆Bεx0¯. Furthermore, dx0,0=x02<ε, that is, 0∈Bεx0¯.Now, we prove the Kannan type fixed point theorem for a completeCRbMS, which extends the results in the paper [3], replacing Picard’s iteration algorithm by Mann’s iteration algorithm.Theorem 22.
LetX,d,w be a CRbMS with constant s≥1 and the mapping T:X⟶X be defined by
(51)dTx,Ty≤kdx,Tx+dy,Ty,for all x,y∈X, and k∈0,1/2. Let the sequence xn generated by the Mann iterative process and x0∈X such that dx0,Tx0=M<∞. If k∈0,1/3s and αn∈0,1/us2 (u is an arbitrary real number and u>5), then T has a unique fixed point in X. Moreover, the sequence xn⊆Bεx0¯ and xn⟶x∗∈Bεx0¯ as n⟶∞, if the following inequality holds:
(52)dx0,Tx0≤k1−skε.Proof.
Without loss of generality, we suppose thatxn≠xn+1 for all n≥ℕ. Indeed, If xn=xn+1, that is, xn=wxn,Txn;αn. Then, we have
(53)dxn,Txn=dwxn,Txn;αn,Txn≤αndxn,Txn,and it shows that dxn,Txn=0 and xn=Txn, which means that xn is a fixed point of T, and the proof is finished. Thanks to Definition 5 and Definition 6, we have
(54)dxn,xn+1=dxn,wxn,Txn;αn≤1−αndxn,Txn.Now, we have the following two cases:Case 23.
Ifxn≠Txn−1 for all n∈ℕ, we have
(55)dxn,Txn=dwxn−1,Txn−1;αn−1,Txn≤αn−1dxn−1,Txn+1−αn−1dTxn−1,Txn≤sαn−1dxn−1,xn+dxn,Txn−1+dTxn−1,Txn+1−αn−1dTxn−1,Txn≤sαn−11−αn−1dxn−1,Txn−1+αn−1dxn−1,Txn−1+kdxn−1,Txn−1+kdxn,Txn+1−αn−1kdxn−1,Txn−1+dxn,Txn≤sαn−1+sαn−1k+1−αn−1kdxn−1,Txn−1+sαn−1k+1−αn−1kdxn,Txn,which establishes that
(56)1−1−αn−1k−sαn−1kdxn,Txn≤sαn−1+sαn−1k+1−αn−1kdxn−1,Txn−1.
Notice that1−αn−1k+sαn−1k<1, then we have
(57)dxn,Txn≤sαn−1+sαn−1k+1−αn−1k1−1−αn−1k−sαn−1kdxn−1,Txn−1.
Sinceu>5, we conclude that
(58)sαn−1+sαn−1k+1−αn−1k1−sαn−1k−1−αn−1k≤1/us+1/3us2+1/3s1−1/3us2−1/3s≤u+42u−1<1;Case 24.
Ifxn=Txn−1 for some n∈ℕ, then
(59)dxn,Txn=dTxn−1,Txn≤kdxn−1,Txn−1+dxn,Txn,and this implies that
(60)1−kdxn,Txn≤kdxn−1,Txn−1.
Since0≤k≤1/3s, then we get
(61)dxn,Txn≤k1−kdxn−1,Txn−1.
Noticing that(62)k1−k≤12≤u+42u−1.Letλu=u+4/2u−1, it is clear that λu<1, and for any n∈ℕ, we obtain the following inequality:
(63)dxn,Txn≤λudxn−1,Txn−1,foralln∈ℕ,and it implies that dxn,Txn is a decreasing sequence of nonnegative reals. Hence, there exists γ≥0 such that
(64)limn⟶∞dxn,Txn=γ.We will show thatγ=0. Suppose that γ>0. Letting n⟶∞ in inequality (63), we obtain
(65)γ≤λuγ,a contradiction. Hence, we get that γ=0. Moreover, we have
(66)dxn,xn+1≤1−αndxn,Txn,which shows that limn⟶∞dxn,xn+1=0. Next, we shall prove that limn⟶∞dxn,xn+2=0. In order to do it, we will consider the following two cases:Case 25.
ifxn+1≠Txn for all n∈ℕ, then we obtain
(67)dxn,xn+2≤dxn,wxn+1,Txn+1;αn+1≤αn+1dxn,xn+1+1−αn+1dxn,Txn+1≤αn+1dxn,xn+1+1−αn+1sdxn,Txn+dTxn,xn+1+dxn+1,Txn+1≤αn+1dxn,xn+1+1−αn+1sdxn,Txn+αndTxn,xn+dxn+1,Txn+1.
Hence,(68)dxn,xn+2≤αn+1dxn,xn+1+1−αn+1s1+αndTxn,xn+dxn+1,Txn+1.Case 26.
If there exist somen∈ℕ such that xn+1=Txn, then we get
(69)dxn,xn+2≤sdxn,xn+1+dxn+1,xn+3+dxn+3,xn+2=sdxn,xn+1+dxn+1,wxn+2,Txn+2;αn+2+dxn+3,xn+2≤sdxn,xn+1+αn+2dxn+1,xn+2+1−αn+2dxn+1,Txn+2+1−αn+2dTxn+2,xn+2≤sdxn,xn+1+αn+2dxn+1,xn+2+1−αn+2kdxn,Txn+1−αn+2kdxn+2,Txn+2+1−αn+2dTxn+2,xn+2.
Hence,(70)dxn,xn+2≤sdxn,xn+1+αn+2dxn+1,xn+2+1−αn+2kdxn,Txn+1−αn+2kdxn+2,Txn+2+1−αn+2dTxn+2,xn+2.
It follows from (68) and (70) that
(71)limn⟶∞dxn,xn+2=0.
Next, we will claim thatxn is a Cauchy sequence by contradiction. Assume there exists ε0>0 and the subsequences xθk and xηk of xn such for θk>ηk>k with dxθk,xηk≥ε0, dxθk−1,xηk<ε0. On the one hand,
(72)ε0≤dxθk,xηk≤sdxθk,xηk+1+dxηk+1,xηk+2+dxηk+2,xηk,taking the limit superior in above inequality as k⟶∞, and we get
(73)ε0s≤limsupk⟶∞dxθk,xηk+1.
On the other hand, letxηk≠xηk+2≠xθk−1≠xηk+1 and Txηk≠xηk≠Txθk−1≠xηk+1, and we have
(74)dxθk,xηk+1=dwxθk−1,Txθk−1;αθk−1,xηk+1≤αθk−1dxθk−1,xηk+1+1−αθk−1dTxθk−1,xηk+1≤αθk−1sdxθk−1,xηk+dxηk,xηk+2+dxηk+2,xηk+1+1−αθk−1sdTxθk−1,Txηk+dTxηk,xηk+dxηk,xηk+1≤αθk−1sdxθk−1,xηk+dxηk,xηk+2+dxηk+2,xηk+1+1−αθk−1skdxθk−1,Txθk−1+kdxηk,Txηk+dTxηk,xηk+dxηk,xηk+1.We obtain
(75)ε0s≤limsupk⟶∞dxθk,xηk+1≤1usε0<1sε0,a contradiction. Thus, xn is a Cauchy sequence in X. Since the space X,d,w is complete, there exists x∗∈X such that limn⟶∞xn=x∗. We shall show that x∗ is a fixed point of T. Applying the rectangular inequality, we obtain that
(76)dx∗,Tx∗≤sdx∗,xn+dxn,xn+1+dxn+1,Tx∗≤sdx∗,xn+sdxn,xn+1+sαndxn,Tx∗+1−αndTxn,Tx∗≤sdx∗,xn+sdxn,xn+1+s2αndxn,xn+1+dxn+1,x∗+dx∗,Tx∗+s1−αnkdxn,Txn+kdx∗,Tx∗,since s2αn+s1−αnk<1, and then
(77)dx∗,Tx∗≤11−s2αn−s1−αnksdx∗,xn+sdxn,xn+1+s2αndxn,xn+1+dxn+1,x∗+s1−αnkdxn,Txn,letting n⟶∞, and we deduce dx∗,Tx∗=0 which implies Tx∗=x∗. Thus, x∗ is a fixed point of T. Suppose that x∗,y∗∈X are two distinct fixed points of T, that is, Tx∗=x∗,Ty∗=y∗. Then,
(78)0<dx∗,y∗=dTx∗,Ty∗≤kdx∗,Tx∗+dy∗,Ty∗=0,which is a contradiction. Therefore, we must have dx∗,y∗=0, that is, x∗=y∗. Thus, T has a unique fixed point. Finally, we will prove the iteration sequence xn⊆Bεx0. In order to complete it, we will use mathematical induction. Choose x0∈X, and we have
(79)dx0,x1=dx0,wx0,Tx0;α0≤1−α0dx0,Tx0≤1−α0β1−sβε<ε,which implies dx0,x1<ε; therefore, x1∈Bεx0. Suppose x2,x3,⋯,xm∈Bεx0. It is easy to see that sk1−sk<2/9s. Without loss of generality, we can assume that xm+1≠xm. If Tx0≠Txm≠x0≠xm+1, then
(80)dx0,xm+1≤sdx0,Tx0+dTx0,Txm+dTxm,xm+1≤sk1−skε+k21−skε+kλumdx0,Tx0+αnλumdTx0,x0≤sk1−skε+k21−skε+λumk21−sk+αmλmk1−skε≤2ε9+2ε27+2ε27+2ε9<ε.We also need to distinguish the following four cases:Case 27.
Ifx0=Tx0, then we have
(81)dx0,xm+1=dx0,wxm,Txm;αm≤αmdx0,xm+1−αmdTx0,Txm≤αmε+1−αmkdx0,Tx0+dxm,Txm≤αmε+2k1−sk1−αmkε<ε.Case 28.
Ifx0=Txm, then we have
(82)dx0,xm+1=dTxm,wxm,Txm;αm≤αmdTxm,xm≤αmλumdTx0,x0≤αmλumk1−skε<ε.Case 29.
Ifxm+1=Tx0, then we have
(83)dx0,xm+1=dx0,Tx0≤k1−skε<ε.Case 30.
Ifxm+1=Txm, then we have
(84)dx0,xm+1=dx0,wxm,Txm;αm≤αmdx0,xm+1−αmdx0,Txm=αmdx0,xm+1−αmdx0,xm+1,which implies
(85)dx0,xm+1≤dx0,xm≤ε.Finally, by above cases, we prove thatdx0,xm+1≤ε, which show that xm+1∈Bεx0¯. Hence, by induction xn∈Bεx0¯. Therefore, we conclude that xn∈Bεx0¯ for all n∈ℕ. As every closed ball in a complete metric space is complete, so x∗∈Bεx0¯, as n⟶∞.Next, we give the following example to illustrate above theorem.Example 31.
LetX=ℝ+∪0 and the mapping T:X⟶X such that
(86)Tx=0,ifx∈0,2,12x,ifx∈2,+∞,for any x,y∈X. Let us define the metric d:X×X⟶X by the formula dx,y=x−y2 as well as the mapping w:X×X×0,1⟶X by the formula wx,y;α=αx+1−αy. Choose x0≥0 to be the initial value and xn+1=wxn,Txn;αn, where αn=1/49. If k=1/9, then xn∈Bεx0¯, and T has a unique fixed point in Bεx0¯.Proof.
It is easy to see thatX,d,w is a CRbMS with s=3. We claim that T satisfies inequality
(87)dTx,Ty≤kdx,Tx+dy,Ty,for any x,y∈X. Next, we will consider the four cases:
(a)
Ifx,y∈0,2, then it is easy to see that inequality (87) holds(b)
Ifx∈0,2 and y∈2,+∞, then(88)dTx,Ty−19dx,Tx+dy,Ty=12y2−19x2+y−12y2≤12y2−19y−12y2≤0,which implies that
(89)dTx,Ty≤19dx,Tx+dy,Ty,holds for any x∈0,2 and y∈2,+∞.
(c)
Ifx∈2,+∞ and y∈0,2, then, similarly to case (b), we can also get that inequality (87) holds(d)
Ifx,y∈2,+∞, then(90)dTx,Ty−19dx,Tx+dy,Ty=141x−1y2−19x−12x2+y−12y2=141x2+1y2−2xy−19x2+y2+14x2+14y2−2=8361x2+1y2−19x2+y2+12xy−2≤83612+12−192+2+14−2=836−936<0,which shows that
(91)dTx,Ty<19dx,Tx+dy,Tyholds for all x,y∈2,+∞.
Summarizing, inequality, (87) holds for all x,y∈X. Next, we will show that T has a unique fixed point in X. In order to do it, we will consider the following two cases:
(i)
Ifx0<2, then(92)Tx0=0,x1=149x0+4849Tx0=149x0,Tx1=0,x2=149x1+4849Tx1=1492x0,Tx2=0,⋯xn=149xn−1+4849Txn−1=149nx0.
Obviously,xn⟶0 as n⟶∞,
(ii)
Ifx0≥2, then(93)Tx0=12x0,x1=149x0+4849Tx0,x1x0=149+4849×12x02≤1349.
If0≤x1<2, then Tx1=0. From the case (i), it follows that xn⟶0 as n⟶∞. If x1≥2, then x2/x1=1/49+48/49×1/2x12≤13/49. From the above procedure, without loss of generality, we can assume that xn−1≥2. Then, we obtain
(94)xnxn−1=149+4849×12x12≤1349,xnx0=x1x0×x2x1×⋯×xnxn−1≤1349n,which implies that xn≤13/49nx0.
Hence,limn⟶∞xn=0, where 0 is a fixed point of T. Actually, 0 is a unique fixed point of T in . Indeed, suppose that y∗∈2,+∞ is a fixed point of T, then Ty∗=y∗, that is, y∗=Ty∗=1/2y∗, which implies y∗=2/2<2, a contradiction. Thus, T has a unique fixed point in . Let ε=x02/k1−sk>0, then k1−skε=x02≥dx0,Tx0. For all n∈ℕ, from above proof, we can obtain xn≤13/49nx0, then dx0,xn=x0−13/49nx02<x02<ε, and this means that the sequence xn⊆Bεx0¯. Furthermore, dx0,0=x02<ε, that is, 0∈Bεx0¯, and the proof is finished.The concept of well posedness is very important in many fields of mathematics and has evoked much interest to several researchers [27–29].Definition 32. (see [26]).
LetX,d be a metric space and T be a self-map. The fixed point problem of T is said to be well posed if
T has a unique fixed point x∗∈X(2)
For any sequenceyn in X with limn⟶∞dyn,Tyn=0, we have limn⟶∞dyn,x∗=0We next study the well posedness of the fixed point problem ofT in complete CRbMS.Theorem 33.
LetX,d,w be a CRbMS with constant s≥1 and all the hypotheses of Theorem 12 hold. If the constant 0<α<1, then fixed point problem of T is well posed.Proof.
Letx∗ is a unique fixed point of T and assume yn be a sequence in X such that limn⟶∞dyn,Tyn=0. Because of uniqueness of the fixed point of T, for all n∈ℕ, we can assume that yn≠Tyn. If yn=wyn,Tyn;α for some α∈0,1, n∈ℕ, then
(95)dyn,x∗=dwyn,Tyn;α,x∗≤αdyn,x∗+1−αdTyn,x∗≤α+1−αβdyn,x∗,since α+1−αβ<1, and we get dyn,x∗=0. Due to α>0, it is not difficult to see that Tyn≠wyn,Tyn;α, indeed, if not,
(96)dyn,Tyn=dyn,wxn,Tyn;α≤1−αdyn,Tyn,a contradiction. Therefore, let us assume that yn≠Tyn≠wyn,Tyn;α, and then
(97)dyn,x∗≤sdyn,wyn,Tyn;α+dwyn,Tyn;α,Tyn+dTyn,x∗≤s1−αdyn,Tyn+sαdyn,Tyn+sβdyn,x∗,combining with 1−sβ>0, and we obtain
(98)dyn,x∗≤s1−sβdyn,Tyn,which implies limn⟶∞dyn,x∗=0, which completes the proof.Theorem 34.
LetX,d,w be a CRbMS with constant s≥1 and all the hypotheses of Theorem 22 hold. If the constant 0<α<1, then fixed point problem of T is well posed.Proof.
Letx∗ be a unique fixed point of T and a sequence yn in sequence in X such that limn⟶∞dyn,Tyn=0. Without loss of generality, let yn≠x∗, for all n∈ℕ. By the help of uniqueness of the fixed point of T, then we have yn≠Tyn. If yn=wyn,Tyn;α for some α∈0,1, n∈ℕ, then
(99)dyn,x∗=dwyn,Tyn;α,x∗≤αdyn,x∗+1−αdTyn,x∗≤αdyn,x∗+1−αkdyn,Tyn+dx∗,Tx∗≤αdyn,x∗+1−αkdyn,Tyn.
Hence,(100)dyn,x∗≤kdyn,Tyn,and we conclude that limn⟶∞dyn,x∗=0. Due to α>0, it is not difficult to see that Tyn≠wyn,Tyn;α, indeed, if not,
(101)dyn,Tyn=dyn,wyn,Tyn;α≤1−αdyn,Tyn,a contradiction. Therefore, let us assume that yn≠Tyn≠wyn,Tyn;α, and then
(102)dyn,x∗≤sdyn,wyn,Tyn;α+dwyn,Tyn;α,Tyn+dTyn,x∗≤s1−αdyn,Tyn+sαdyn,Tyn+skdyn,Tyn+dx∗,Tx∗,combining with 1−sβ>0, and we obtain
(103)dyn,x∗≤s+skdyn,Tyn,which implies limn⟶∞dyn,x∗=0, which completes the proof.


## 3. Applications

In this section, we will apply our result to solving the following functional equation arising in dynamic programming:(104)pu=supv∈Bfu,v+Gu,v,pφu,v,for all u∈A, where f:A×B⟶R, φ:A×B⟶A, and G: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. Precisely, see also [30, 31]. Let X=RA denote the set of all bounded real-valued functions on A and the norm · defined as x=supu∈Axu for all x∈X. Clearly, X,· is a Banach space. Moreover, we can define a rectangular b−metric d by
(105)dx,y=supu∈Axu−yu2,for all x,y∈X. Since X,· is complete, we deduce that X,d is a complete rectangular b−metric space with s=3. In order to show the existence of a solution of equation (104), we consider the operator T:X⟶X of the form
(106)Txu=supv∈Bfu,v+Gu,v,xφu,v,for all u∈A and x∈X. We will prove the following theorem.Theorem 35.
LetT:X⟶X be given by (106). Suppose that the following hypotheses hold:
(A1)f:A×B⟶R and G:A×B×R⟶R are bounded functions;
(A2) There existsa>0, for all u∈A, v∈B and x,y∈X, such that
(107)G1u,v,xu−G2u,v,yu≤axu−yu.Then, the functional equation (104) has a bounded solution.Proof.
Obviously,T is well defined, since f and G are bounded. That is, Tx∈X and operator T are well defined. Then, from (A2), we have
(108)Txu−Tyu2=supv∈Bfu,v+Gu,v,xφu,v−supv∈Bfu,v+Gu,v,yφu,v2≤supv∈BGu,v,xφu,v−Gu,v,yφu,v2≤a2supv∈Bxu−yu2.
Let0≤a≤1/32; thus, all the conditions of Theorem 12 are fulfilled, and there exists a fixed point x∗∈X of T such that Tx∗=x∗. In other words,
(109)x∗u=supv∈Bfu,v+Gu,v,x∗φu,v,for all u∈A. This completes the proof.Example 36.
Consider the functional equation(110)xu=supv∈0,1sinu+v+ln1+uv+16xuvfor u∈0,2. We let A=0,2, B=0,1. f:A×B⟶R is defined by fu,v=sinu+v, φ:A×B⟶A is defined by φu,v=uv, and
(111)G:A×B×R⟶Ris defined by Gu,v,x=ln1+uv+1/6x for x∈X. It is not difficult to see that f and G are bounded functions. Moreover,
(112)Gu,v,xφu,v−Gu,v,yφu,v2=ln1+uv+x−ln1+uv+x2=ln1+uv+1/6x+1/6y−1/6y1+uv+1/6y2=ln1+1/6x−1/6y1+uv+1/6y2≤ln1+16x−16y2≤136x−y2.Thus, all the conditions of Theorem35 are fulfilled. Hence, functional equation (110) has a solution x∗u∈RA.



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