# New Results for Set-Valued Mappings on Ordered Metric Spaces

Authors: F. Sabetghadam; H. P. Masiha; H. Aydi
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We establish some new fixed point results for order-closed multivalued mappings in complete metric spaces endowed with a partial order.


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## Body

## 1. Introduction

In 1976, an order relation was defined in metric spaces [1]. Later, many researchers proved various fixed point results in this setting (see [2–11]), while the authors in [4, 5, 9] considered coupled questions for so-called monotone conditions. Zhang [12] proved some (coupled) fixed point theorems for multivalued mappings with monotone conditions in metric spaces with a partial order. Since then, Agarwal and Khamsi [13] extended Caristi’s fixed point to vector-valued metric spaces. Also, Chung [14] considered nonlinear contraction mappings. Nadler [15] defined multivalued contractions, and Assad and Kirk [16] proved fixed point theorems for set-valued mappings (see also the related works [17–22]).In this paper, we present some fixed point results in ordered metric spaces for order-closed multivalued operators. First, we need some facts.Lemma 1.
[23] Let E,d be a metric space and ϕ:E⟶−∞,+∞ be a functional. Consider γ:0,+∞⟶0,+∞ a nondecreasing, continuous, and subadditive function so that γ−10=0. Take the relation “≼” on E given as
(1)u≼viffγdu,v≤ϕu−ϕv,forallu,v∈E.
Then, “≼” is a partial order relation on E. Apparently, u≼v then ϕu≥ϕv.Here, we state some definitions. LetE be a topological space. Denote by NE the family of nonempty subsets of E. Let ≼ be a partial order on E.Definition 2.
[24] Given two nonempty subsets UandV of E.
(2)r1Ifforeacha∈U,thereisb∈Vsothata≼b,thenU≺1Vr2Ifforeachb∈V,thereisa∈Usothata≼b,thenU≺2Vr3IfU≺1VandU≺2V,thenU≺V.Note that≺1 and ≺2 are different relations between U and V (see Remark 114 of [24]). Also, ≺1, ≺2, and ≺ are not partial orders on NE (see Remark 2 of [24]).Definition 3.
Ifun⊂E satisfies u1≼u2≼⋯≼un≼⋯ or u1≽u2≽⋯≽un≽⋯, then un is called a monotone sequence.Definition 4.
A multivalued operatorG:E⟶NE is said to be order-closed if, for monotone sequences un and vn⊂E, we have un⟶u0, vn⟶v0, and vn∈Gun imply v0∈Gu0.Definition 5.
A functiong:E⟶−∞,+∞ is said to be order upper (lower) semicontinuous, if, for a monotone sequence un⊂E with u0∈E, we have
(3)un⟶u0⇒limn→∞¯gun≤gu0,gu0≤limn→∞¯gun.Note that an upper (lower) semicontinuous function is an order upper (lower) semicontinuous. But the converse is not true (see Remark 3 of [24]).


## 2. Main Results

LetE,d be a metric space. For ϕ:E⟶ℝ, define the partial order “≼” on E induced by ϕ and γ in Lemma 1.Theorem 6.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be a bounded below function. Suppose that G:E⟶NE is order-closed with respect to “≼” so that
(4)Hforanyu∈E,thereisv∈Gusothatu≼v
Then, there is a monotone sequenceunn=0∞⊂E, so that un+1∈Gun for n=0,1,2,⋯, and un converges to u∗, which is a fixed point of G. If in addition, ϕ is order lower semicontinuous on E, then un≼u∗ for each n.Proof.
By the conditionH, take u0∈E. From H, there is u1∈Gu0 so that u0≼u1. Again, from H, there is u2∈Gu1 with u1≼u2. Continuing this procedure, we have an increasing sequence un so that un+1∈Gun. Therefore,
(5)ϕun−1≥ϕunforalln.
That is, the real sequenceϕun is decreasing, so it is a convergent sequence because ϕ is bounded from below on E.
Using Remark 3 of [25] and Remark 2 of [26], we find that
(6)limℓ→0+γℓℓ=supγℓℓ:ℓ>0,provided that γ is subadditive. Thus,
(7)liminfℓ→0+γℓℓ>0.
By (7), there are δ>0 and c>0 so that
(8)γℓ≥cℓ,forallℓ∈0,δ.
Sinceγ is nondecreasing, we have γℓ≥γδ for each ℓ∈δ,+∞.
Let0<ε<γδ. Then
(9)γℓ>ε,foreachℓ∈δ,+∞,i.e.,
(10)ifγℓ≤ε,thenℓ∈0,δ.
Therefore, we have(11)ℓ≥0:γℓ≤ε⊂0,δ,which together with (8) implies that
(12)γℓ≥cℓ,forallℓ∈ℓ≥0:γℓ≤ε.
Note thatϕun is convergent; then, there is n0 so that for all m≥n≥n0,
(13)γdun,um≤ϕun−ϕum<ε.
Moreover, by (12), we get
(14)cdun,um≤γdun,um≤ϕun−ϕum,forallm≥n≥n0.
The convergence ofϕun implies that un is a Cauchy sequence. By the completeness of E, there is u∗∈E so that un⟶u∗ as n⟶∞. Since G is order-closed, un is monotone and un+1∈Gun. Consequently,
(15)u∗∈Gu∗,i.e., u∗ is a fixed point of G.
Ifϕ is order lower semicontinuous on E, by definition of “≼,” then we have for each m(16)γdum,u∗=limn→∞γdum,un≤limn→∞¯ϕum−ϕun=ϕum−limn→∞¯ϕun≤ϕum−ϕu∗.
It yields thatum≼u∗. The proof is completed.The proof of the following theorem carries over in the same manner as for Theorem6.Theorem 7.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be a bounded below function. Suppose that G:E⟶NE is order-closed with respect to “≼.” Assume that
(17)Hforeachu∈E,thereisv∈Gusothatv≼u
Then, there is a monotone sequenceunn=0∞⊂E, un+1∈Gun, n=0,1,2,⋯, such that un converges to u∗, which is a fixed point of G. If, in addition, ϕ is order upper semicontinuous on E, then u∗≼un for all n.Theorem 8.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be bounded below function. Suppose that G:E⟶NE is order-closed with respect to “≼.” Assume that
(i)
for eachu,v∈E with u≼v⇒Gu≺1Gv(ii)
there existsu0∈E such that, u0≺1Gu0.
Then,G has a fixed point u∗, and there is a sequence un with un−1≼un∈Gun−1 for n≥1, such that un⟶x∗. Moreover, if ϕ is order lower semicontinuous, then un≼u∗ for all n.Proof.
SinceGu≠∅, by ii, we can choose u1∈Gu0 so that x0≼u1. This implies that Gu0≺1Gu1, by definition of ≺1, there is u2≼Gu1 so that u1≼u2. Continuing this procedure, we can find an increasing sequence un such that un∈Gun−1. The rest of the proof is similar as in Theorem 6.The following supports Theorem8.Example 9.
LetE=0,1 and dξ,σ=∣ξ−σ∣, for ξ,σ∈E. The metric space E,d is complete. Consider the multivalued mapping G:E⟶NE given as
(18)Gξ=ξ5,ξ4forξ∈E∩ℚξ5=3,ξ2forξ∈Eℚ,where Q is the set of rational numbers. Let γs=ss≥0 and ϕξ=∣ξ∣ξ∈E. Note that ϕ is bounded from below. Take the order ≼ induced by ϕ, that is, given as follows:
(19)ξ≼σ⇔γdξ,σ≤ϕξ−ϕσ.
Mention thatG verifies the following assertions:
(a)
for eachu,v∈E with u≼v, we have Gu≺1Gv(b)
0≺1G0(c)
G is order-closed on E.
Hence,G has a fixed point on E, which is u∗=0.Theorem 10.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be a bounded below function. Suppose G:E⟶NE is order-closed with respect to “≼.” If the following conditions hold:
(i)
for eachu,v∈E with u≼v⇒Gu≺2Gv(ii)
there existsu0∈E such that, Gu0≺2u0
thenG has a fixed point u∗. Also, there is a sequence un with un−1≽un∈Gun−1 for any n≥1 such that un⟶u∗. Moreover, if ϕ is order upper semicontinuous, then un≽u∗ for all n.
The following example illustrates Theorem10.Example 11.
LetE=0,1, dξ,σ=∣ξ−σ∣, ϕξ=−ξ for ξ,σ∈E and γt=t for each t≥0. Here, E is a complete metric space, and ϕ is a bounded below function. Consider the order ≼ induced by ϕ:
(20)ξ≼σ⇔dξ,σ≤ϕξ−ϕσ.
Clearly, this partial order is the usual order onE. Define G:E⟶NE by Gξ=ξ/3,ξ/2. It is obvious that G satisfies the following:
(a)
for eachu,v∈E with u≼v⇒Gu≺2Gv(b)
G1≺21(c)
G is order-closed on E.
Hence,G has a fixed point on E, which is u∗=0.Now, a multivalued version of Theorem 2 of [27] may be obtained. Here, the considered multivalued mapping is not necessarily continuous.Theorem 12.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be a continuous function bounded below. Let G:E⟶NE be a multivalued mapping. Suppose that
(21)H1foranyu∈Ethereisv∈Gusothatu≼vH2Guiscompactforeachu∈E.
Then,G has a fixed point.Proof.
Set(22)M≔u∈E∣∃v∈Gu:v≽u.
We claim thatM has a maximum element. For a directed set I, let uii∈I be a totally ordered subset in M. For i,j∈I with i≤j, the fact that ui≼uj yields ϕui≥ϕuj. Due to the fact that ϕ is bounded below, ϕui is a convergent set in ℝ. Consider
(23)γdui,uj≤ϕui−ϕuj.
As in proof of Theorem6, ui is Cauchy in E, which is complete, so ui converges to u in E. For j∈I,
(24)γduj,u=limiγduj,ui≤limiϕuj−ϕui=ϕuj−ϕu.
Hence,uj≼u for each j∈I. By H1, for each vj∈Guj, there is wj∈Gu so that vj≼wj. By the compactness of Gu, there is a convergence subset vj′ of vj. Assume that vj′ converges to v∈Gu. Take I′ such that i′≤j′ implies vj≼wj≼wj′. One writes
(25)γdvj,v=limj′γdvj,wj′≤limj′ϕvj−ϕwj′=ϕvj−ϕv.
Sovj≼v for all j. Also,
(26)γdu,v=limjγdvj,v≤limjϕvj−ϕv=ϕu−ϕv.
Sou≼v and u∈M. Thus, uj has an upper bound in M.
By Zorn’s Lemma, there is a maximum elementu∗∈M. also, there is v∗∈Gu∗ so that u∗≼v∗. Using H1, there is z∗∈Gv∗ so that v∗≼z∗. Hence, v∗∈M. The element u∗ is maximum in M, so v∗=u∗ and u∗∈Gu∗. That is, u∗ is a fixed point of G.Theorem 13.
LetE,d,≼ be a complete ordered metric space and ϕ:E⟶−∞,+∞ be a continuous function bounded below. Let G:E⟶NE be a multivalued mapping. Assume that
(27)H1Foranyu∈E,thereisv∈Gusothatv≼uH2Guiscompactforanyu∈E.
Then,G has a fixed point.Remark 14.
IfG is a continuous single-valued mapping in Theorem 6 (resp., Theorem 7), we can replace the condition H by
(28)H′:foreachu∈E,u≼Guresp.,Gu≼u,and we can obtain the same result.Remark 15.
If in Theorem8 (resp., Theorem 10), G is assumed to be a continuous single-valued mapping, then we get the same result when replacing the conditions i and ii by
(i)
G is monotone increasing (resp., decreasing), that is, for u≼v, we have Gu≼Gvresp.,Gv≼Gu(ii)
there existsu0 with u0≼Gu0 (resp., Gu0≼u0).
If in additionϕ is order upper (resp., lower) semicontinuous on E, then u∗ is the smallest (resp., largest) fixed point of G in K=u∈E∣ssu0≼u (resp., K=u∈E∣u≼u0).Proof.
Letz be a fixed point of G in K, i.e., z=Gz. Since u0≼z, we get Gu0≼Gz. Hence, u1=Gu0≼Gz=z, i.e., u1≼z. Suppose un≼z, then Gun≼Gz, i.e., un+1=Gun≼Gz=z. By a mathematical induction, we have un≼z for all n, then
(29)γdu∗,z=limn→∞dun,z≤limn→∞¯ϕun−ϕz=limn→∞¯ϕun−ϕz≤ϕu∗−ϕz.
That is,u∗≼z.
Note that if we omit the continuity ofG and add above condition on ϕ, the results remain true.



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*Source: 5393521-2020-09-24.xml*