# Best Proximity Point Theorems for Cyclic Contractions Mappings in Banach Algebras

Authors: K. Chaira; S. Chaira; S. Lazaiz
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this paper, we present some new best proximity point theorems for three operators acting in Banach algebras. An application is given to show the usefulness and the applicability of the obtained results.


---

## Body

## 1. Introduction

The study of functional integral equations and differential equations is the main object of research in nonlinear functional analysis. These equations occur in physical, biological, and economic problems. Some of these equations can be formulated into nonlinear operator equations:(NOE)x=Ax·Bx+Cxin suitable Banach algebras.Recently, many authors are interested on the study of equation (NOE) and obtained some interesting results (see for instance [1–5]). In 2010, Ben Amar et al. [6] proved some existence fixed point theorems which allowed them to solve equation (NOE) where the involved operators are weakly sequentially continuous.LetX be a Banach algebra with a norm ∥·∥. Let Ω1,Ω2 be a pair of nonempty subsets of X. Given two mappings A and C defined on X and an operator B:Ω1∪Ω2⟶X. Under suitable conditions, we define the operator T:=I−C/A−1∘B:Ω1∪Ω2⟶Ω1∪Ω2 such that TΩ1⊂Ω2 and TΩ2⊂Ω1. If Ω1∩Ω2 is nonempty, then the mapping T restricted to Ω1∩Ω2 is a self mapping. Then, a solution of equation (NOE) is a fixed point of T. Furthermore, if the fixed point equation x=Ax.Bx+Cx does not possess a solution it is natural to explore to find an x∗∈Ω1 satisfying(1)x∗−Tx∗=distΩ1,Ω2,where distΩ1,Ω2=infx−y:x∈Ω1,y∈Ω2. This point x∗∈Ω1 is said to be the best proximity point of T. Note that a point x∈Ω1∪Ω2 is the best proximity point of T if x is a solution of the minimization problem(2)minx∈Ω1∪Ω2dx,Tx.The best proximity point notion can be viewed as a generalization of fixed point, since most fixed point theorems can be derived as corollaries of the best proximity point theorems.The first result of this kind is due to Fan (see [7], Theorem 2) which is stated in normed spaces for continuous mappings. In [8], Eldred and Veeramani introduced the concept of cyclic contraction mappings and gave the best proximity point results for this class of mappings. They also gave an algorithm to reach this best proximity point where the space is uniformly convex. Furthermore, in [9], Taghafi and Shahzad proved the existence of the best proximity point for a cyclic contraction mapping in a reflexive Banach space.For noncyclic mappings, i.e.,TΩ1⊂Ω1 and TΩ2⊂Ω2, Gabelah and Künzi in [10] established some best proximity point results in the framework of complete CAT(0) spaces. In addition, they gave an approach to reach this best proximity point by means of an algorithm. Regarding the relationship between the noncyclic and cyclic results, the authors in [11] proved that the existence of best proximity points for cyclic nonexpansive mappings is equivalent to the existence of best proximity pairs for noncyclic nonexpansive mappings in the setting of strictly convex Banach spaces. For more on the best proximity point results, the interested reader can consult [12–18].The paper is organized as follows. After some preliminaries, in Section 3, we prove the existence of the best proximity point where the involving operators areD-Lipschitzs and cyclic contraction (see Theorem 1). Also, an example is given to illustrate the obtained result. In Theorem 1, we consider the case where X is a uniformly convex Banach algebra. In Section 4, we show the applicability of our result (Theorem 1) to the theory of nonlinear integral equations:(FIS1)xt=Kt,xt+Txt·q2+∫0tgs,ysds(FIS2)yt=Kt,yt+Tyt·q1+∫0tfs,xsds,where Ω1 and Ω2 two subsets of the Banach algebra E=CJ,X of all continuous functions from J to X.


## 2. Preliminaries

Definition 1.
An algebraX is a vector space endowed with an internal composition law noted by · i.e.,(3)·:X×X⟶Xx,y⟶x·ywhich is associative and bilinear.
A normed algebra is an algebra endowed with a norm satisfying the following property for allx,y∈X; x·y≤xy. A complete normed algebra is called a Banach algebra.Definition 2.
LetX be a Banach space with norm ∥·∥. A mapping T:X⟶X is called D-Lipschitz if there exists a continuous nondecreasing function ΦT:ℝ+⟶ℝ+ satisfying(4)Tx−Ty≤ΦTx−yfor all x,y∈X with ΦT0=0. In the special case when ΦTr=αr for some α>0, T is called lipschitzian mapping with a Lipschitz constant α.Definition 3.
LetX,· be a Banach space. We say that X is uniformly convex if for every ε>0,(5)δε=inf1−x+y2:x≤1,y≤1andx−y≥ε>0,
The functionδ is known as the modulus of uniform convexity of X. Note that any uniformly convex Banach space is reflexive.Theorem 4 (see [17]).
LetX be a uniformly convex Banach space. Let Ω1 be a nonempty closed bounded convex subset of X such that Ω10 is compact, and Ω2 be a nonempty closed convex subset of X. Let T:Ω1∪Ω2⟶Ω1∪Ω2 be a relatively nonexpansive mapping. Then, there exists x∗∈Ω1 such that x∗−Tx∗=distΩ1,Ω2.The authors in [8] introduced the following notion of cyclic contraction.Definition 5.
LetΩ1 and Ω2 be nonempty subsets of a metric space X. A mapping T:Ω1∪Ω2⟶Ω1∪Ω2 is said to be a cyclic contraction if it satisfies:(1)
TΩ1⊂Ω2 and TΩ2⊂Ω1(2)
for somek∈0,1, dTx,Ty≤kdx,y+1−kdistΩ1,Ω2, for all x∈Ω1, y∈Ω2
SincedistΩ1,Ω2≤dx,y, for x∈Ω1 and y∈Ω1, dTx,Ty≤dx,y for all x∈Ω1y∈Ω2, i.e., T is relatively nonexpansive.
We conclude this section by recalling some best proximity point results for this class of mappings.Theorem 6 (see [8]).
LetΩ1 and Ω2 be nonempty closed subsets of a complete metric space X. Let T:Ω1∪Ω2⟶Ω1∪Ω2 be a cyclic contraction mapping and x0∈Ω1. Define xn+1=Txn, n∈ℕ. Suppose x2n:n∈ℕ has a convergent subsequence in Ω1, then there exists x∈Ω1 such that dx,Tx=distΩ1,Ω2.Theorem 7 (see [8, 19]).
LetΩ1 and Ω2 be nonempty closed convex subsets of a uniformly convex Banach space. Suppose T:Ω1∪Ω2⟶Ω1∪Ω2 is a cyclic contraction mapping. Then, T has a unique best proximity point in Ω1. Further, if x0∈Ω1 and xn+1=Txn, then the sequence x2nn≥0 converges to the best proximity point.


## 3. Main Results

We start this section by introducing the notion ofα,β-monotone property for a pair of functions.Definition 8.
Letα,β∈ℝ+2 and ϕ,ψ:ℝ+⟶ℝ+ be two mappings. We say that the pair ϕ,ψ has the property α,β-monotone if(i)
ϕ0=0=ψ0,(ii)
I−α·ϕ−β·ψ is nondecreasing on ℝ+ and limr→+∞r−αϕr−βψr=+∞Remark 9.
Ifϕ,ψ has the property α,β-monotone and ϕ,ψ are continuous, then the mapping I−α·ϕ−β·ψ:ℝ+⟶ℝ+ is invertible.Example 1.
Letϕ,ψ:ℝ+⟶ℝ+ be the mappings defined by:(6)ϕr=rr+1andψr=r2,forallr∈ℝ+.
So, the mappingI−1/2ϕ−ψ:ℝ+⟶ℝ+ has the property 1/2,1-monotone.
Recall that an operatorA from a Banach algebra X is said to be regular on X if A maps X into the set of all invertible elements of X.Theorem 10.
LetΩ1,Ω2 be a nonempty closed pair of a Banach algebra X. Let A,C:X⟶X and B:Ω1∪Ω2⟶X be three operators which satisfy the following conditions:(1)
A is regular on X and A<1(2)
A and C are D-Lipschitzs with the D-functions ΦA and ΦC, respectively, BΩ1∪Ω2 is bounded with bound M, and MΦAr+ΦCr≤1−Ar for all r>0, and I−MΦA−ΦC is nondecreasing(3)
B is cyclic contraction mapping on Ω1∪Ω2(4)
suppose there exists a sequencexnn≥0 of Ω1∪Ω2 such that x0∈Ω1, Bxn=I−C/Axn+1 and the sequence x2nn≥0 has a convergent subsequence in Ω1,(5)
y=Ay·Bx+Cy,x∈Ω1⇒y∈Ω2x=Ax·By+Cx,y∈Ω2⇒x∈Ω1
Then, there existsx,y∈Ω1×Ω2 such that(7)x−Ax.Bx−CxAx=distΩ1,Ω2=y−Ay·By−CyAy.Proof.
Lety be fixed in Ω1∪Ω2 and let us define the mapping Fy on X by(8)Fyx=Ax·By+Cx,forallx∈X.
Letx1,x2∈X. The use of assumption ii leads to(9)Fyx1−Fyx2≤Ax1·By−Ax2·By+Cx1−Cx2≤Ax1−Ax2By+Cx1−Cx2≤MΦAx1−x2+ΦCx1−x2
Now an application of Boyd and Wong’s fixed point theorem [20], Theorem 1 leads to the existence of a unique point xy∈X such that Fyxy=xy. Hence, the operator T:=I−C/A−1B:Ω1∪Ω2⟶X is well defined.
Moreover, assumptionv implies that TΩ1⊂Ω2 and TΩ2⊂Ω1. Indeed, let x∈Ω1 and y∈X such that y=Ay.Bx+Cy, so Tx=I−C/A−1Bx=y∈Ω2. Similarly, for all y∈Ω2, Ty∈Ω1. Hence, T is cyclic on Ω1∪Ω2.
T is cyclic contraction on Ω1∪Ω2. Indeed, let x,y∈Ω1×Ω2, the use of assumption ii and iii and the fact that Tz=ATz·Bz+CTz for all z∈Ω1∪Ω2 leads to(10)Tx−Ty≤ATx·Bx−ATy·By+CTx−CTy≤ATx−ATyB+Bx−ByA+CTx−CTy≤Akx−y+1−kdistΩ1,Ω2+MΦATx−Ty+ΦCTx−Ty
SinceΦA,ΦC has the property M,1-monotone, we have(11)Tx−Ty≤I−MΦA−ΦC−1Akx−y+1−kdistA,B≤kx−y+1−kdistΩ1,Ω2.
Byiv, there exists a sequence xnn≥0 of Ω1∪Ω2 such that x0∈Ω1, Txn=xn+1 and the sequence x2nn≥0 has a convergent subsequence in Ω1.
Thus, by Theorem6, there exists x1,y1∈Ω1×Ω2 such that x1−Tx1=distΩ1,Ω2=y1−Ty1.
Lety=I−C/A−1Bx1∈Ω2. By iii, B is cyclic contraction on Ω1∪Ω2, By∈Ω1and I−C/Ay∈Ω2, so(12)distΩ1,Ω2≤y−Ay·By−CyAy=I−CAy−By≤Bx1−By≤x1−y=x1−I−CA−1Bx1=distΩ1,Ω2.
Similarly,x−Ax·Bx−Cx/Ax=distΩ1,Ω2, where x=I−C/A−1By1∈Ω1.Example 2.
LetX=ℝ endowed with the usual norm ∣·∣ and let Ω1=1/6,1/4, Ω2=3/8,1.(i)
LetB the function defined on Ω1∪Ω2 by(13)Bx=38ifx∈Ω114ifx∈Ω2.
Letx,y∈Ω1×Ω2, we have x−y=y−x≥3/8−1/4=1/8.(14)Bx−By=38−14=18=distΩ1,Ω2≤kdistΩ1,Ω2+1−kdistΩ1,Ω2≤kx−y+1−kdistΩ1,Ω2,wherek∈0,1.
Thus,B is cyclic contraction on Ω1∪Ω2 and M=BΩ1∪Ω2=3/8.(ii)
LetA the function defined on ℝ by Ax=1/3, for all x∈ℝ. The function A is D-Lipschitz with the D-function ΦA=0, and A=1/3(iii)
LetC the function defined on X by(15)Cx=16ifx∈X\Ω214ifx∈Ω2.
For eachx,y∈Ω1×Ω2,(16)Cx−Cy=16−14=112=23·18≤23·x−y.
The functionC is D-Lipschitz with the D-function defined by ΦCr=2/3.r, for all r∈ℝ+. We have I−MΦA−ΦC=I−ΦC:r↦2/3r is nondecreasing, and for all r>0,(17)MΦA+ΦCr=23r≤1−Ar.(iv)
Letx∈ℝ and y∈Ω2. Suppose x=Ax·By+Cx and y=Ay·Bx+Cy. We have(18)x=Ax·By+Cx=13·14+16=14∈Ω1,y=Ay·Bx+Cy=13·38+14=38∈Ω2.(v)
For ally∈Ω2, I−C/A−1y=y/3+1/4, so for each x∈Ω1, I−C/A−1Bx=I−C/A−13/8=3/8. Thus, for any sequence xnn≥0 of Ω1∪Ω2 such that x0∈Ω1, Bxn=I−C/Axn+1, the sequence x2nn≥0 has a convergent subsequence
Hence, by Theorem1, there exists x,y∈Ω1×Ω2 such that(19)x−Ax.Bx−CxAx=distΩ1,Ω2=y−Ay·By−CyAy.where x,y=1/4,3/8.Theorem 11.
LetΩ1,Ω2 be a nonempty closed pair of a Banach algebra X. Let B:Ω1∪Ω2⟶X and C:X⟶X be two operators which satisfy the following conditions:(1)
C is D-Lipschitz with the D-function ΦC, ΦCr<r, for all r>0(2)
B is cyclic contraction mapping on Ω1∪Ω2(3)
I−C−1 is relatively nonexpansive mapping on Ω1∪Ω2(4)
suppose there exists a sequencexnn≥0 of Ω1∪Ω2 such that x0∈Ω1, Bxn=I−Cxn+1 and the sequence x2nn≥0 has a convergent subsequence in Ω1(5)
y=Bx+Cy,x∈Ω1⇒y∈Ω2x=By+Cx,y∈Ω2⇒x∈Ω1
Then, there existsx,y∈Ω1×Ω2 such that(20)x−Bx−Cx=distΩ1,Ω2=y−By−Cy.Proof.
Byi, we show that T:=I−C−1·B:Ω1∪Ω2⟶X is well defined. Moreover, the use of assumptions ii, iii, and v shows that T is cyclic contraction on Ω1∪Ω2. Indeed, by v, it is cyclic. Let x,y∈Ω1×Ω2, the use of assumptions ii and iii leads to Bx,By∈Ω2×Ω1,(21)Tx−Ty=I−C−1Bx−I−C−1By≤Bx−By≤kx−y+1−kdistΩ1,Ω2.
Byiv, there exists a sequence xnn≥0 of Ω1∪Ω2 such that x0∈Ω1, Txn=xn+1 and the sequence x2nn≥0 has a convergent subsequence in Ω1.
Thus, by Theorem6, there exists x1,y1∈Ω1×Ω2 such that(22)x1−Tx1=distΩ1,Ω2=y1−Ty1.
Lety=I−C−1Bx1∈Ω2. We have By∈Ω1 and I−Cy∈Ω2. By iii, B is cyclic contraction on Ω1∪Ω2, so(23)distΩ1,Ω2≤y−By−Cy=I−Cy−By≤Bx1−By≤kx1−y+1−kdistΩ1,Ω2≤x1−y=x1−I−C−1Bx1=distΩ1,Ω2
Similarly, we getx−Ax·Bx−Cx=distΩ1,Ω2, where x=I−C−1By1∈Ω1.Remark 12.
Under the same hypotheses of the previous theorem whereC=0, we obtain the classical result. That is, there exists x,y∈Ω1×Ω2 such that(24)x−Bx=distΩ1,Ω2=y−By.Theorem 13.
LetΩ1,Ω2 be a nonempty closed convex pair of a uniformly convex Banach algebra X. Let A,C:X⟶X and B:Ω1∪Ω2⟶X be three operators which satisfy the following conditions:(1)
A is regular on X and A<1(2)
A and C are D-Lipschitzs with the D-functions ΦA and ΦC, respectively, BΩ1∪Ω2 is bounded with bound M, and MΦAr+ΦCr≤1−∥A∥r for all r>0 and I−MΦA−ΦC is nondecreasing(3)
B is cyclic contraction on Ω1∪Ω2(4)
y=Ay·Bx+Cy,x∈Ω1⇒y∈Ω2x=Ax·By+Cx,y∈Ω2⇒x∈Ω1.
Then, there exists a uniquex,y∈Ω1×Ω2 such that(25)x−Ax.Bx−CxAx=distΩ1,Ω2=y−Ay·By−CyAy.
Further, ifx0∈Ω1 and Bxn=I−C/Axn+1, n∈ℕ, then the sequence x2nn≥0 converges to the best proximity point.Proof.
Byi, ii, iii, and v, we show that T:=I−C/A−1·B:Ω1∪Ω2⟶X is well defined and cyclic contraction on Ω1∪Ω2.
Thus, by Theorem7 and iv, there exists a unique x,y∈A×B such that(26)x−Ax·Bx−CxAx=distΩ1,Ω2=y−Ay·By−CyAy,and if x0∈Ω1 with Bxn=I−C/Axn+1, i.e., Txn=xn+1 for all n∈ℕ, then the sequence x2nn≥0 converges to the best proximity point.


## 4. Application

LetE=CJ,ℝ the Banach algebra of all continuous functions from J=0,1 to ℝ, endowed with the sup-norm ∥·∥∞, defined by(27)x∞=supxt,t∈0,1,for each x∈CJ,ℝ. Let q1,q2∈ℝ2 and suppose q1<0<q2.We consider the closed and nonempty sets(28)Ω1=x∈E:xt≥q2,∀t∈JΩ2=y∈E:yt≤q1,∀t∈J.For anyx,y∈Ω1×Ω2 and for all t∈J, we have(29)x−y∞≥xt−yt=xt−yt≥q1−q2,so distΩ1,Ω2=q1−q2. We consider the following two nonlinear functional integral equations(FIS1)xt=Kt,xt+Txt·q2+∫0tgs,ysds(FIS2)yt=Kt,yt+Tyt·q1+∫0tfs,xsds,where x,y∈Ω1×Ω2 and t∈0,1.The integral equations (FIS1)–(FIS2) may be written, respectively, as:(30)xt=Axt.Byt+Cxtyt=Ayt.Bxt+Cyt,where x,y∈Ω1×Ω2 and t∈J. To simplify the notations, we put(31)Axt=Txt,Bxt=q1+∫0tfs,xsdsifx∈Ω1q2+∫0tgs,xsdsifx∈Ω2,Cxt=Kt,xt.The goal of this section is to apply our main result to investigate the existence of an optimum solutionx,y of the (FIS1)–(FIS2) problem in the sense that the pair x,y satisfies:(32)x−Ax·Bx−CxAx=y−Ay·By−CyAy=distΩ1,Ω2.Note that, ifx is a solution of (FIS1) and y is a solution of (FIS2), then the pair x,y need not form an optimum solution see [21], pp 27-31 for more details.We consider the following assumptions:(i)
(a)
The functionK·,x·:J×ℝ⟶ℝ is continuous, for all x∈E(b)
There is a continuous functionδ:J⟶0,+∞ with bound Δ=supt∈Jδt such that Kt,xt−Kt,yt≤δtxt−yt, for all x,y∈E and t∈0,1(ii)
(a)
The functionsf,g:J×ℝ⟶ℝ are measurables and(33)N=maxsupx∈Ef·,x·∞,supx∈Eg·,x·∞<∞(b)
Letx,y∈Ω1×Ω2. For all s∈J,(34)fs,xs−gs,ys≤αxs−ys−q1−q2(c)
For alls∈J, fs,ys≤0≤gs,xs(iii)
(a)
T:E⟶E is D-Lipschitz with the D-function ΦT, such that ΦT is differentiable on ℝ+ and ΦA′<1−Δ/Q+N, N≠0, where Q=maxq1,q2(b)
T is regular on CJ,ℝ, and T=supx∈ETx∞<1(iv)
The familyBx:x∈Ω1∪Ω2 is equicontinuous and closedTheorem 14.
Assume the hypothesesH1−H4 hold. ΦTr≤1−∥T∥−Δ/Q+Nr for all r>0 and Tx>0 for all x∈E. Moreover,(35)K·,x·q2ifx∈Ω1K·,x·q1ifx∈Ω2.
Then, there exists an optimum solutionx,y∈Ω1×Ω2 for (FIS1)–(FIS2) problem.Proof.
(1)(i)
ByH1, we have K·,x·−K·,y·∞≤Δx−y∞ for all x,y∈E, so Cx−Cy∞≤ΦCx−y∞ for all x,y∈E, where ΦCr=Δr, for all r≥0(ii)
-Letx∈Ω1, We have, for all t∈J(36)Bxt=q1+∫0tfs,xsds≤Q+N.
Similarly, for ally∈Ω2 and t∈J we get Byt≤Q+N. Hence, M=BΩ1∪Ω2≤Q+N<∞.(iii)
-Furthermore, by hypothesis we haveΦAr≤1−∥A∥−Δ/Q+Nr, for all r>0. So(37)MΦAr+ΦCr=MΦAr+Δr≤Q+NΦAr+Δr≤1−Ar,forallr>0.(iv)
We show thatI−MΦA−ΦC is nondecreasing. Let r,r′∈ℝ+ such that r<r′.
Since,ϕA is nondecreasing and differentiable on ℝ+, with ∣ΦA′∣<1−Δ/Q+N, so(38)0≤ΦAr′−ΦArr′−r<1−ΔQ+N≤1−ΔM,thus,(39)I−MΦA−ΦCr<I−MΦA−ΦCr′.
That is,I−MΦA−ΦC is nondecreasing.
(2)(i)
Lett,t′∈J such that t<t′,(40)Bxt−Bxt′=∫tt′fs,xss≤∫tt′fs,xsds≤Nt−t′
Thus,Bx is Lipschitzian, so Bx∈E. Let x∈Ω1 and t∈J. By H2, iii, We have(41)Bxt=q1+∫0tfs,xsds≤q1.
Hence,Bx∈Ω2. Similarly, we get BΩ2⊂Ω1.(ii)
-Letx,y∈Ω2×Ω1 and t∈J(42)Bxt−Byt=q1−q2+∫0tfs,xs−gs,ysds.
ByH2, iii, we have fs,xs−gs,ys≤0, for all s∈J, so(43)q1−q2+∫J2∩0,tfs,xs−gs,ysds≤0.
Then,(44)Bxt−Byt=q2−q1+∫0tgs,xs−fs,ysds≤q1−q2+∫0tαxs−ys−q1−q2ds≤q1−q2+∫0tαx−y∞−q1−q2ds≤1−αq1−q2+αx−y∞.
Thus,Bx−By∞≤αx−y∞+1−αdistΩ1,Ω2 which shows that B is cyclic contraction.
(3) Lety∈Ω2 and x∈E such that x=Ax·By+Cx. We show that x∈Ω1. We have, for all t∈J, Byt≥0, Axt≥0 and Cxt≥q2, so(45)xt=Axt.Byt+Cxt≥Cxt≥q2.
Letx∈Ω1 and y∈E such that y=Ay·Bx+Cy. We show that y∈Ω2. We have, for all t∈J, Byt≤0, Axt≥0 and Cxt≤q1, so(46)yt=Ayt.Bxt+Cyt≤Cxt≤q1.
(4) AsBΩ1∪Ω2<∞, so the family Bx:x∈Ω1∪Ω2 is uniformly bounded; by H4, this family is equicontinuous. Therefore, by Arzela-Ascoli’s theorem, Bx:x∈Ω1∪Ω2 lies in a compact subset of Ω1∪Ω2. Let xnn≥0 be a sequence of Ω1∪Ω2 such that I−C/A−1Bxn=xn+1, i.e., Bxn=I−C/Axn+1. We have Bx2n−1n≥1 has a convergent subsequence Bx2σn−1n≥1. Let u=limn→+∞Bx2σn−1. As BΩ1∪Ω2 is closed, there exists z∈Ω1∪Ω2 such that Bz=u. We obtain, for each n∈ℕ∗(47)Tx2σn−1−Tz∞≤ATx2σn−1·Bx2σn−1−ATz·Bz∞+CTx2σn−1−CTz∞≤ATx2σn−1−ATz∞B+Bx2σn−1−Bz∞A+CTx2σn−1−CTz∞≤ABx2σn−1−Bz∞+MΦATx2σn−1−Tz∞+ΦCTx2σn−1−Tz∞.
Thus,(48)Tx2σn−1−Tz∞≤I−MΦA−ΦC−1ABx2σn−1−Bz∞≤Bx2σn−1−Bz∞.
Hence,(49)x2σn−Tz∞≤Bx2σn−1−Bz∞.
Which prove that the sequencex2σnn≥0 is convergent.
Thus, by Theorem1, there exists x,y∈Ω1×Ω2 such that(50)x−Ax·Bx−CxAx=distΩ1,Ω2=y−Ay·By−CyAy.



---

*Source: 8889508-2020-11-05.xml*