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

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. Let E 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.

If un⊂E satisfies u1≼u2≼⋯≼un≼⋯ or u1≽u2≽⋯≽un≽⋯, then un is called a monotone sequence.

Definition 4.

A multivalued operator G: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 function g: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

Let E,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.

Let E,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 sequence unn=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 condition H, 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 sequence ϕun 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.

Since γ is nondecreasing, we have γℓ≥γδ for each ℓ∈δ,+∞.

Let 0<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 that ϕun 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 of ϕun 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.

If ϕ 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 that um≼u∗. The proof is completed.

The proof of the following theorem carries over in the same manner as for Theorem 6.

Theorem 7.

Let E,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 sequence unn=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.

Let E,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

for each u,v∈E with u≼v⇒Gu≺1Gv

there exists u0∈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.

Since Gu≠∅, 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 Theorem 8.

Example 9.

Let E=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 that G verifies the following assertions:

for each u,v∈E with u≼v, we have Gu≺1Gv

0≺1G0

G is order-closed on E.

Hence, G has a fixed point on E, which is u∗=0.

Theorem 10.

Let E,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:

for each u,v∈E with u≼v⇒Gu≺2Gv

there exists u0∈E such that, Gu0≺2u0

then G 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 Theorem 10.

Example 11.

Let E=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 on E. Define G:E⟶NE by Gξ=ξ/3,ξ/2. It is obvious that G satisfies the following:

for each u,v∈E with u≼v⇒Gu≺2Gv

G1≺21

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.

Let E,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 that M 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 Theorem 6, 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.

So vj≼v for all j. Also,
(26)γdu,v=limjγdvj,v≤limjϕvj−ϕv=ϕu−ϕv.

So u≼v and u∈M. Thus, uj has an upper bound in M.

By Zorn’s Lemma, there is a maximum element u∗∈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.

Let E,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.

If G 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 Theorem 8 (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

G is monotone increasing (resp., decreasing), that is, for u≼v, we have Gu≼Gvresp.,Gv≼Gu

there exists u0 with u0≼Gu0 (resp., Gu0≼u0).

If in addition ϕ 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.

Let z 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 of G and add above condition on ϕ, the results remain true.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no competing interests regarding the publication of this paper.

Authors’ Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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