# A Fixed-Point Theorem for Ordered Contraction-Type Decreasing Operators in Banach Space with Lattice Structure

Authors: Chenguang Wang; Jinxiu Mao; Zengqin Zhao
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this work, we mainly improve the results in Amini-Harandi and Emami (2010). By introducing a new kind of ordered contraction-type decreasing operator in Banach space, we obtain a unique fixed point by using the iterative algorithm. An example is also presented to illustrate the theorem.


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## Body

## 1. Introduction

In this work, we obtain a unique fixed point for a kind of ordered contraction-type decreasing operator in Banach space by using the iterative algorithm. The fixed-point study is mainly focused on two aspects. On the one hand, it is about the research of contraction-type mapping, for example, in [1–8]. On the other hand, it is about the study of monotone operators with concavity and convexity, for example, in [9–15]. There is little research on operators that only satisfy the partial-order constrictions. Applications of operator theory in fractional differential equations can be seen in [16–43].The following generalization of Banach’s contraction principle is due to Geraghty [44].Lemma 1.
LetM,d be a complete metric space and let f:M⟶M be a map. Suppose there exists β∈ζ such that for each x,y∈M,
(1)dfx,fy≤βdx,ydx,y,where ζ denotes the class of those functions β:0,∞⟶0,1 which satisfy the condition βtn⟶1⇒tn⟶0. Then f has a unique fixed-point z∈M, and fnx converges to z, for each x∈M.Very recently, Amini-Harandi and Emami [6] proved the following existence theorem which is a version of Lemma 1 in the context of partially ordered complete metric spaces:Lemma 2.
Let (M,≼) be a partially ordered set and suppose that there exists a metric d in M such that M,d is a complete metric space. Let f:M⟶M be an increasing map such that there exists an element x0∈M with x0≼fx0. Suppose that there exists β∈ζ such that
(2)dfx,fy≤βdx,ydx,y,x,y∈M,y≤x.
Assume that eitherf is continuous or M is such that if an increasing sequence xn⟶x in M, then xn≼x,∀n.
Besides, if for eachx,y∈M, there exists z∈M, which is comparable to x and y.
Thenf has a unique fixed point.We found that in (2), the contraction is concerning a metric. But in fact, the relation of partial order does not play any role in (2). That is to say, in (1), the constriction y≼x is not effective because dx,y=dy,x. A question appears naturally in the authors’ minds: “Can the contraction condition be merely about partial order so that y≼x can be effective?” The authors have been haunted by this question since it was found. Driven by this idea, we introduce a new kind of ordered contraction-type decreasing operator in Banach space with lattice structure and obtain a unique fixed point of the operator. Our results are helpful and meaningful for studies of fixed point. Comparing to [6], our improvements are in three aspects.First, the contraction is merely about partial order, and the relation of partial-ordery≼x does play an important role in the contraction condition. This has never been seen. Second, we consider the situation when the operator is decreasing. Third, we only use the iterative algorithm, and we can start the iterative process with any initial point, i.e., we do not need any assumptions of the existence of upper or lower solutions. An example is also presented to illustrate the theorem.The outline of this paper is as follows. In the remainder of this section, we will give some preliminaries. In Section2 of this paper, we present the existence and uniqueness theorem. In Section 3, an example is illustrated.Definition 3 (see [45]).
LetE be a real Banach space. A nonempty convex closed set P⊂E is called a cone if
(i)
x∈P,λ≥0⇒λx∈P(ii)
x∈P,−x∈P⇒x=θ; θ is the zero element in EIn the case thatP is a given cone in a real Banach space E,., a partial order “≤” can be induced on E by x≤y⇔y−x∈P. The cone P is called normal if there exists a constant N>0, such that for all x,y∈E,θ≤x≤y implies that x≤Ny. Details about cones and fixed point of operators can be found in [45, 46].Definition 4 (see [47, 48]).
We call a setX⊂E a lattice under the partial ordering ≤, if supx,y and infx,y exist for arbitrary x,y∈X.Lemma 5 (see [45]).
A coneP is normal if and only if there exists a norm .1 in E which is equivalent to . such that for any θ≤x≤y,x1≤y1,i.e.,.1 is monotone. The equivalence of .and.1 means that there exist M>m>0 such that m.1≤.≤M.1.Lemma 6 (see [45]).
LetP be a normal cone in a real Banach space E. Suppose that xn is a monotone sequence which has a subsequence xni converging to x∗, then xn also converges to x∗. Moreover, if xn is an increasing sequence, then xn≤x∗n=1,2,3,⋯; if xn is a decreasing sequence, then x∗≤xnn=1,2,3,⋯.


## 2. The Main Results

We suppose thatE is a partially ordered Banach space. P is a normal cone. The partial-order “≤” on E is induced by the cone P.Letζ denote the class of those functionals β:P⟶0,1 which satisfy the condition
(3)βwn⟶1⇒wn⟶θforanymonotonicsequencewn⊂PTheorem 7 (main theorem).
Suppose thatX⊂E is a closed subset, P⊂X. X is a lattice. A:X⟶X is a decreasing operator and satisfies the following ordered contraction condition:
(H) Suppose that there exists β∈ζ such that
(4)Au−Av≤βv−uv−u,∀u,v∈X,u≤v.
ThenA has unique fixed-point u∗∈X. Moreover, constructing successively sequence
(5)un=Aun−1n=1,2,⋯,for any initial u0∈X, we have
(6)limn→∞un=u∗.Remark 8.
Here, we study the decreasing operator while most of the contractions are about increasing operators. And the contraction condition (4) is merely about the partial order, while most of the contractions are about metric.Remark 9.
Two elementsx and y in an ordered set X,≤ are said to be comparable if either x≤yory≤x, and we denote it as x≤≥y.Proof.
Letu0∈X, we have Au0∈X. So we have the following two cases.Case 1.
Whenu0 is comparable with Au0. Firstly, without loss of generality, we suppose that
(7)u0≤Au0.
Ifu0=Au0, then the proof is finished. Suppose that u0<Au0. Since A is decreasing, we obtain Au0≥A2u0, and it is easy to prove that A2 is increasing. Using the contractive condition (4), we have
(8)Au0−A2u0≤βAu0−u0Au0−u0≤Au0−u0,hence,
(9)u0≤A2u0.From (4), we have
(10)A2v−A2u≤βAu−AvAu−Av≤βAu−Avβv−uv−u≤βv−uv−u,∀u,v∈X,u≤v.Let(11)Bu=A2u,∀u∈X.From (9) and (10), we have the following two conclusions:
(a)
There exists a functionalβ∈ζ such that for u,v∈X with u≤v(12)Bv−Bu≤βv−uv−u(b)
There existsu0∈X such that u0≤Bu0We assert that the operator B has unique fixed point in X. And the unique fixed point of B is also the unique fixed point of A. In order to be clear, we divide the process of proof into three steps.Step 1.
We will use the method of iteration to construct a fixed point ofB. In fact, consider the iterative sequence
(13)un+1=Bun,n=0,1,2,⋯.Sinceu0≤Bu0 and the operator B is increasing, we have
(14)u0≤u1≤⋯≤un≤⋯.This means thatun is an increasing sequence. It follows from (12) that
(15)θ≤un+1−un=Bun−Bun−1≤βun−un−1un−un−1.SinceP is normal, from the equivalence of .and.1 in Lemma 5 we have
(16)un+1−un1=Bun−Bun−11≤βun−un−1un−un−11.Then,un+1−un1 is a decreasing sequence and bounded as follows. So
(17)limn→∞un+1−un1=r≥0.Assumer>0. Then, from (16), we have
(18)un+1−un1un−un−1∣1≤βun−un−1.The above inequality yields(19)limn→∞βun−un−1=1.Andβ∈ζ implies limn→∞un−un−1=θ. Then,
(20)limn→∞un−un−11=0and r=0.Now we show thatun is a Cauchy sequence. On the contrary, assume that
(21)limm,n→∞supun−um1>0.For any fixed natural numbern,mn>m. From (16), by the triangle inequality un−um1≤un−un+11+un+1−um+11+um+1−um1≤un−un+11+βun−umun−um1+um+1−um1.Hence, we haveun−um1≤1−βun−um−1un−un+11+um+1−um1.Sincelimsupm,n→∞un−um1>0 and limn→∞un−un+11=0, then
(22)limsupm,n→∞1−βun−um−1=∞,from which we obtain
(23)limsupm,n→∞βun−um=1.But sinceβ∈ζ, we get
(24)limsupm,n→∞un−um1=0.This contradicts (21) and shows that un is a Cauchy sequence in X.SinceX is closed, we can suppose that there exists a u∗∈X such that
(25)un⟶u∗.SinceP is normal, (14) together with Lemma 5 implies that
(26)un≤u∗.(26), together with (12) and the equivalence of .1 and ., implies that
(27)Bu∗−Bun1≤βu∗−unu∗−un1.Sou∗−Bu∗1≤u∗−un+1∣1+Bu∗−un+11=u∗−un+1∣1+Bu∗−Bun1≤u∗−un+11+βu∗−unu∗−un1.Letn⟶∞, we obtain u∗−Bu∗1=0. Sou∗=Bu∗.This proves thatu∗ is a fixed point of B in X and
(28)u∗=limn→∞Bnu0.Step 2.
We will obtain the uniqueness of the fixed point ofB. On the contrary, if u¯ is another fixed point of B, we will get u¯=u∗.In fact, the first case, whenu¯is comparable withu0, without loss of generality, we suppose thatu¯≤u0. Since B is increasing,
(29)u¯=Bnu¯≤Bnu0.Moreover, by (12)
(30)Bnu0−u¯=Bnu0−Bnu¯≤βBn−1u0−Bn−1u¯Bn−1u0−Bn−1u¯=βBn−1u0−Bn−1u¯Bn−1u0−u¯,and so
(31)Bnu0−u¯1≤βBn−1u0−u¯Bn−1u0−u¯1.Consequently, the sequence(32)γn=Bnu0−u¯1,is nonnegative and decreasing, and so limn→∞γn=γ≥0. Now we show that γ=0.On the contrary, assume thatγ>0. By passing to subsequences, if necessary, we may assume that limn→∞βBnu0−u¯=λ exists. From (31), we obtain γ≤λγ, and so λ=1. Since β∈ζ, then limn→∞Bnu0−u¯=θ, and γ=limn→∞Bnu0−u¯1=0.This contradiction provesγ=0.It can be obtained that(33)u¯=limn→∞Bnu0=u∗.The second case, whenu¯ cannot compare with u0. FromXwhich is a lattice, we obtain
(34)y1=infu¯,u0∈X,satisfying
(35)y1≤u¯,y1≤u0,i.e., u¯ is comparable with y1 and u0 is comparable with y1. Since B is increasing, we know
(36)Bny1≤Bnu¯,Bny1≤Bnu0,n=1,2,⋯.Moreover, by (12)
(37)Bnu0−Bny1≤βBn−1u0−Bn−1y1Bn−1u0−Bn−1y1.So we have
(38)Bnu0−Bny11≤βBn−1u0−Bn−1y1Bn−1u0−Bn−1y11≤Bn−1u0−Bn−1y11.Similar to the process of (31), (32), and (33),
(39)limn→∞Bny1=limn→∞Bnu0=u∗.From (36), we have
(40)limn→∞Bny1=limn→∞Bnu¯=u¯.So from (39) and (40), we get
(41)u¯=u∗.(33) together with (41) implies that u∗ is unique fixed point of B.Step 3.
We will point that the unique fixed point ofB is also the unique fixed point of A. Since
(42)A2u∗=Bu∗=u∗.Thus,(43)A2Au∗=AA2u∗=Au∗,i.e., BAu∗=Au∗. From the uniqueness of the fixed point of B, we know
(44)Au∗=u∗.Sou∗ is the unique fixed point of A in X.Case 2.
Another case, whenu0 is not comparable to Au0. From X which is a lattice, we know there exists v0∈E such that infAu0,u0=v0. That is, v0≤Au0,v0≤u0. Since A is a decreasing operator, we have
(45)Au0≤Av0.
This shows that(46)v0≤Av0.
For any initialv0∈X, constructing successively sequence
(47)vn=Avn−1.
FromB=A2, we can get
(48)v2n=A2nv0=Bnv0,n=1,2,⋯.
From (28), we know
(49)limn→∞v2n=limn→∞Bnv0=u∗.
Sincev2n+1=A2n+1v0=BnAv0 and from the arbitrary of v0 in (28), we obtain
(50)limn→∞v2n+1=limn→∞BnAv0=u∗.(49) and (50) imply that
(51)limn→∞vn=u∗,holds. Similarly to the proof of Step 2 and Step 3 in case 1, we get that u∗ is the unique fixed point of A.


## 3. An Example

LetE=R, equipped with usual normal ·=∣·∣ and usual partial order ≤. X=0,∞,P=0,∞. Then, P⊂X⊂E. P is a normal cone in X. X,·,≤ is a partially ordered Banach space. And X is a lattice under the partial order ≤ induced by the cone P.Then,X=0,∞,P=0,∞ satisfying the assumptions of Theorem 7. Define the mapping A:X⟶X by Ax=m+1+mx/1+x,x∈X where m is a fixed real number. Then, A is nonincreasing. Define β:P⟶0,∞ by βw=1/1+w, then β∈ζ. Now, for all x,y∈X with x≤y, we have
(52)Ax−Ay=m+1+mx1+x−m+1+my1+y=y−x1+x+y+xy≤y−x1+y−x=βy−xy−x,so that A and β satisfy the assumption of Theorem 7. Observing that all the other conditions of Theorem 7 are also satisfied, A has a unique fixed-point x∗>0. Moreover, constructing successively sequence
(53)xn+1=m+1+mxn1+xnn=0,1,2,⋯,for any initial x0≥0, we have
(54)limx→∞xn=x∗.



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