R E S E A R C H
Open Access
Fixed point theorems on ordered vector
spaces
Jin Lu Li
1†, Cong Jun Zhang
2†and Qi Qiong Chen
3†*Correspondence:
3Department of Applied
Mathematics, Nanjing University of Science and Technology, Xiaolingwei Street, Nanjing, Jiangsu 210094, China
†Equal contributors
Full list of author information is available at the end of the article
Abstract
In this paper, we introduce the concept of ordered metric spaces with respect to some ordered vector spaces, which is an extension of the normal metric spaces. Then we investigate some properties of ordered metric spaces and provide several fixed point theorems. As applications we prove several existence theorems for best ordered approximation.
MSC: 46B42; 47H10; 58J20; 91A06; 91A10
Keywords: ordered vector spaces; order-continuity; ordered Lipschitz condition; order-preserving mapping; best ordered approximation; fixed point
1 Introduction
LetSbe a nonempty set and letXbe a Banach space. LetKbe a closed convex cone inX. The cone metric defined onS, with respect to the coneKinX, is a bifunctiond:S×S→X, for which the following properties hold:
(a) d(u,v)∈Kandd(u,v) = if and only ifu=v; (b) d(u,v) =d(v,u);
(c) d(u,s) +d(s,v) –d(u,v)∈K, for alls,u,v∈S.
Note that if we takeXto be the set of real numbers and takeK= [,∞), then any cone metric space with respect toKturns out to be a normal metric space. Hence the concept of vector metric spaces extends the notion of the normal metric spaces.
In the field of nonlinear analysis, vector metric spaces have been widely studied by many authors (see []). Similarly to ordinary analysis, the concepts of continuity and of the Lips-chitz condition for mapping on vector metric spaces have been introduced and have been applied for proving the existence of fixed points (see [–]).
Since the definition of vector metrics is based on closed convex cones in normed vector spaces, and every closed convex cone in a vector space can induce a partial order on it, which equips this vector space to be an ordered vector space, it is natural to generalize the concept of cone metric spaces to ordered metric spaces (see Section ).
This paper is organized as follows: in Section , we recall some concepts of ordered vector spaces and order-convergence of sequences; in Section , we introduce the concept of ordered metric spaces and investigate some properties; in Section , several fixed point theorems on ordered metric spaces are provided; in Section , we extend the concept of best ordered approximation and prove several existence theorems by applying the fixed point theorems provided in Section .
2 Preliminaries
In this section, we recall some concepts of ordered sets and some properties of order-limits. For more details, the readers are referred to [–]. Then we extend the concept of order-continuity of mappings on ordered vector spaces and provide some properties that are similar to those of ordinary limits in analysis. These properties will be frequently used throughout this paper.
In this paper, all vector spaces considered are real vector spaces. A vector space X
equipped with a partial order X (that is,X is a reflexive, antisymmetric, and transi-tive binary relation onX) is called a partially ordered vector space, or is simply called an ordered vector space, which is written as (X,X), if the following (order-linearity) prop-erties hold:
(v) xXyimpliesx+zXy+z, for allx,y,z∈X.
(v) xXyimpliesαxXαy, for allx,y∈Xandα≥.
A sequence{xn}in an ordered vector space (X,X) is said to be order-decreasing, which is denoted byxn↓, wheneverm>nimpliesxmXxn. An order-decreasing sequence{xn} is said to order-converge tox, ifxn↓and
{xn}exists with
{xn}=x, which is denoted by{xn} ↓x. The meaning ofxn↑is analogously defined for an order-increasing sequence
{xn}; and{xn} ↑x, if and only ifxn↑and
{xn}exists with
{xn}=x.
Lemma . Let{xn},{yn}be two sequences in an arbitrary ordered vector space(X,X).
The following properties hold:
. xn↓ximpliesaxn↓ax,for everya≥.
. xn↓xandyn↓yimply(xn+yn)↓(x+y).
. xn↓xandyn↓yimply(axn+byn)↓(ax+by),for any nonnegative numbersaandb.
Definition . An ordered vector space (X,X) is said to be generalized Archimedean if and only if for any given elementxX and any decreasing sequence of positive numbers
{an}with limit , we have
anx↓.
LetXbe a vector space andK a closed convex cone inX. Define a binary relationK onXas follows:
xKy if and only if x–y∈K, for everyx,yinX. ()
ThenK is a partial order onX, which satisfies conditions (v) and (v), and therefore (X,K) is an ordered vector space, which is said to be induced by the closed convex coneK.
Lemma . Every ordered vector space induced by a closed convex cone is generalized Archimedean.
The proof of Lemma . is straightforward and it is omitted here.
In order theory, the order completeness of a poset is as important as the topological counterpart of a topological space in analysis. Recall that a subsetCof an ordered vector space (X,X) is said to be chain complete if and only if for any chain{xα}inX,{xα}
Definition . Let (X,X) be an ordered vector space andCa subset ofX.Cis said to be (sequentially) conditionally chain complete if and only if for any sequence{xn}inC, the following properties hold:
. Ifxn↓and{xn}has a lower bound, then
{xn}exists satisfyingxn↓
{xn} ∈C.
. Ifxn↑and{xn}has an upper bound, then
{xn}exists satisfyingxn↑
{xn} ∈C.
3 Ordered metric spaces
We extend the concept of cone metric to the following ordered metric with respect to ordered vector spaces.
Definition . LetSbe a nonempty set and let (X,X) be an ordered vector space. A bi-functiondX:S×S→Xis called an ordered metric onS, with respect toXif, for everyu,
v, andsinS, it satisfies the following conditions:
(m) dX(u,v)XwithdX(u,v) = , if and only ifu=v;
(m) dX(u,v) =dX(v,u);
(m) dX(u,v)XdX(u,s) +dX(s,v).
Then (S,dX) is called an ordered metric space, anddX(u,v) is called the ordered distance betweenuandv, with respect to the ordered vector space (X,X).
We offer some examples below to demonstrate that the class of ordered metric spaces is a very broad one. First, note that (R,≥) is an ordered vector space with the ordinal real order≥. Then we have the following results.
Example . The metric defined on a metric space is an ordered metric; and therefore, every metric space is an ordered metric space.
Example . LetBbe a Banach space with the norm · and letdbe the metric onB
induced by · . Thendis an ordered metric; and therefore, every Banach space with the metric induced by its norm is an ordered metric space (where the metricdis defined by
d(u,v) =u–v, for everyu,v∈B).
Example . Every cone metric on a nonempty set is an ordered metric; and therefore, every cone metric space is an ordered metric space.
As a matter of fact, if an ordered vector space (X,K) is induced by a closed convex cone
K inX, then from (),d(u,v)∈Kif and only ifdX(u,v)K; and therefore, the ordered metrics on sets extend the concept of cone metrics.
Example . Let (X,X) be a Riesz space (vector lattice). Then the bifunctiond X:X×
X→Xinduced by the (ordered) absolute values onXasdX(x,y) =|x–y|, for everyx,y∈X, is an ordered metric onX. Hence every Riesz space with the ordered metric induced by its (ordered) absolute values is an ordered metric space.
Here, as usual, for anyx∈X,x+=x∨,x–= (–x)∨ and|x|=x+∨x–are called the (ordered) positive part, negative part, and (ordered) absolute value ofx, respectively.
space, the ordered distances of pairs of elements are not always comparable. Thus, this idea provides us a useful tool to study some problems, which have outcomes in a poset.
Definition . A sequence {sn} in an ordered metric space (S,dX) is said to order-converge to an elements∈S, which is denoted bysn→–os, whenever there exists another sequence{xn}in (X,X) withxn↓ such that
dX(sn,s)Xxnholds, for eachn. ()
In this case,sis called an order-limit of the sequence{sn}.
Lemma . If a sequence{sn}in an ordered metric space(S,dX)is order-convergent,then
its order-limit is unique.
Proof Letuandvbe order-limits of{sn}. Then there are sequences{xn}and{zn}inXwith
xn↓ andzn↓ such that
dX(sn,u)Xxn and dX(sn,v)Xzn, for eachn.
From condition (m), they imply
dX(u,v)XdX(sn,u) +dX(sn,v)Xxn+zn, for eachn.
From Part of Lemma .,(xn+zn) = . It implies XdX(u,v)X. HencedX(u,v) = . From condition (m) in the definition of ordered metric, it follows thatv=u.
Definition . A sequence{sn} in an ordered metric space (S,dX) is called an order-Cauchy sequence, whenever there exists another sequence {xn} in (X,X) with xn↓ such that
dX(sm,sn)Xxnholds, for eachn, and for everym≥n. ()
Definition . An ordered metric space (S,dX) is said to be order-metric complete, when-ever when-every order-Cauchy sequence inSis order-convergent.
Definition . Let (S,dX) and (T,dY) be ordered metric spaces, with respect to the or-dered vector spaces (X,X) and (Y,Y), respectively. A mappingf :S→T is said to be sequentially orσ-continuous, whenever, for any sequence{sn} ⊂S,sn→–os, with respect to the ordered metricdXonS, impliesf(sn)→–of(s), with respect to the ordered metricdYonT.
4 Several fixed point theorems on ordered metric spaces
Theorem . Let(S,dX)be an order-metric complete ordered metric space,with respect to
a generalized Archimedean ordered vector space(X,X).Let f :S→S be a self-mapping.
Suppose there is a positive numberα< ,such that
dX
f(s),f(s)
Xαd
X(s,s), for all s,s∈S. ()
Proof Taking anys∈S, we define a sequence{fn(s)} ⊂S. From condition (), it follows that
dX
fn+(s),fn(s)
Xαd
X
fn(s),fn–(s)
, for all positive integern.
Iterating the above order-inequality yields
dX
fn+(s),fn(s)
Xαnd
X
f(s),s
, for all positive integern.
By property (m) of order-metric and properties of partial orders in ordered vector spaces, for any positive integeri, we get
dX
fn+i(s),fn(s)
Xαn+id X
f(s),s
+αn+i–dX
f(s),s
+· · ·+αndX
f(s),s
=αnαi+αi–+· · ·+ dX
f(s),s
X αn –αdX
f(s),s
, for all positive integern.
Take
xn=
αn
–αdX
f(s),s
, for all positive integern. ()
Since (X,X) is generalized Archimedean, from Definition ., it yieldsxn↓. It implies that{fn(s)}is an order-Cauchy sequence in (S,dX), which is an order-metric complete ordered metric space. Hence,{fn(s)}has an order-limit, says∗∈C; that is,
fn(s)→–os∗.
Then, from Definition ., there is a sequence{yn} ⊂X, withyn↓ such that
dX
fn(s),s∗
Xy
n, for eachn.
Since the mappingf :S→Shas the ordered Lipschitzian property given by () withxn defined in (), from the above order inequality, it yields
dX
fs∗,s∗XdX
fn+(s),f
s∗+dX
fn+(s),s∗
Xαd X
fn(s),s∗
+dX
fn+(s),s∗
X( +α)y
n, for eachn.
From Lemma .,yn↓ implies ( +α)yn↓. So we obtain
XdX
fs∗,s∗X( +α)yn
= .
It implies
dX
It follows thatf(s∗) =s∗; and therefore,s∗is a fixed point off. The uniqueness of the fixed point off immediately follows from condition () of the mappingf and condition
(m) of the ordered metric.
From Example ., any arbitrary Riesz space (vector lattice) (X,X) is an ordered metric space with the ordered metric induced by its (ordered) absolute values. Then the following result immediately follows from Theorem ..
Corollary . Let S be a nonempty order-metric complete subset of a generalized Archimedean Riesz space.Let f :S→S be a self-mapping.Suppose there is a positive num-berα< ,such that
f(s) –f(s) X|s–s|, for all s,s∈S.
Thenf has a unique fixed point.
Next we generalize the concept of order-increasing of mappings on ordered vector spaces.
(S,dX) and (T,dY) be ordered metric spaces, with respect to ordered vector spaces (X,X) and (Y,Y), respectively. A set-valued mappingF:S→T\∅is said to be ordered metric increasing upward fromSto T\ ∅, whenever there are elementss∈S,t∈Tsuch that, for any givens,s∈S,dX(s,s)XdX(s,s) implies that, for everyt∈F(s), there ist∈F(s) satisfyingdY(t,t)Y dY(t,t) (with respect to pointssandt).Fis said to be ordered metric increasing downward fromSto T\ ∅(with respect to pointssandt), whenever for any givens,s∈S,dX(s,s)XdX(s,s) implies that, for everyt∈F(s), there ist∈F(s) satisfyingdY(t,t)Y dY(t,t).Fis said to be ordered metric increas-ing fromSto T\ ∅(with respect to pointssandt), wheneverFis both ordered metric increasing downward and ordered metric increasing upward. In particular, ifS=T and
s=t, thenFis said to be ordered metric increasing (upward, downward) onS(with re-spect to points).
A single-valued mappingf :S→T is said to be ordered metric increasing (decreas-ing) with respect to elementss∈S,t∈T, whenever, for any givens,s∈S,dX(s,s)X
dX(s,s) impliesdY(f(s),t)Y dY(f(s),t)(dY(f(s),t)YdY(f(s),t)).
Definition . Let (S,dX) be an ordered metric space with respect to an ordered vector space (X,X). (S,d
X) is called a monodromy ordered metric space, with respect to an elements∈S, whenever the mapdX(·,s) :S→Xis one to one. Such an elements∈Sis called a monotonized point of the ordered metric space (S,dX).
Example . For an arbitrary positive integern, let (Rn,n) denote the ordered vector space, wherenis the coordinate ordering on then-dimensional Euclidean spaceRn. LetS be a subset of the positive cone of (Rn,n) containing the origin ofRn. Define an ordered metricdnonSas: for anyx= (x,x, . . . ,xn),y= (y,y, . . . ,yn)∈S,
dn(x,y) =
|x–y|,|x–y|, . . . ,|xn–yn|
.
It follows that
It implies that the ordered metric space (S,dn) is a monodromy ordered metric space with respect to its monotonized point ∈S.
Example . Let (X,X) be a Riesz space and let the ordered metricd
XonXbe defined in Example .. Similarly to Example ., we can see that every subsetSof the positive cone ofXcontaining the origin ofXis a monodromy ordered metric space with the same metricdXonSand with respect to a monotonized point ∈S.
Recall that a nonempty subsetAof a poset (P,) is said to be inductive if every chain in
Ahas an upper bound inA. The next definition extends this concept.
Definition . A nonempty subsetAof a poset (P,) is said to be totally inductive inP
whenever, for any given chain{xα} ⊂P, every elementxβ∈ {xα}has an upper bound inA
implies that the chain{xα}has an upper bound inA.
It is clear that every totally inductive subset of a poset is inductive. However, the inverse may not true. The following lemma provides a sufficient condition for an inductive set to be totally inductive.
Lemma . Let A be an inductive subset of a poset(P,).If A has finite number of maximal elements,then A is totally inductive.
Proof Take an arbitrary chain{xα} ⊂Psatisfying that every elementxβ∈ {xα}has an
up-per boundzβ∈A. We write the set of maximal elements ofAby{u,u, . . . ,um} ⊂A, for some positive integerm. We claim that, for any givenxβ∈ {xα}, we must have
xβzβui, for someiwith ≤i≤m. ()
To prove (), assume, by way of contradiction, that () does not hold for somexγ ∈ {xα}.
Then we define
E(zγ) ={z∈A:zzγ}.
Sincezγ∈E(zγ), we getE(zγ)=∅. For any arbitrary chainCinE(zγ)⊂A, it is also a chain
inA. It follows that the inductivity ofAimplies thatChas an upper boundd∈A. It is clear thatd∈E(zγ). Thus, it shows thatdis an upper bound ofCinE(zγ), henceE(zγ)
is inductive. Using Zorn’s Lemma,E(zγ) has a maximal elementwofE(zγ). In the case if
there isv∈Awithvwandwv, thenv∈E(zγ) andwcould not be a maximal element
inE(zγ). It shows thatwis also a maximal element ofA. From the hypotheses thatzγ does
not satisfy (), then the maximal elementw∈ {/ u,u, . . . ,um}. It is a contradiction to the assumption that{u,u, . . . ,um}is the collection of all maximal elements ofA. The claim is proved.
Following (), a collection of subsets ofCis recursively defined as below:
E=
x∈ {xα}:xu
;
E=
x∈ {xα} \E:xu
E=
x∈ {xα} \(E∪E) :xu
;
· · ·
Em–=
x∈ {xα} \(E∪E· · ·Em–) :xum–
;
Em=
x∈ {xα} \(E∪E· · ·Em–) :xum
.
By (), it follows that{E,E, . . . ,Em}is a partition of{xα}with some empty members. Next we show that in{E,E, . . . ,Em}, there is one and only one nonempty member. To this end, contrarily assume that there are two numbers ≤i<j≤m, such thatEiandEj both are nonempty. Then we takexα(i)∈Eiandxα(j)∈Ej. Since{xα}is a chain, which is totally ordered, so we must have eitherxα(i)xα(j)orxα(j)xα(i). It implies eitherxα(i)∈Ej orxα(j)∈Ei. It is a contradiction to the fact thatEi∩Ej=∅. It proves that in{E,E, . . . ,Em}, there is a unique nonempty member. Thus, this lemma follows immediately.
Theorem . Let(S,dX)be a monodromy ordered metric space with respect to an ordered
vector space(X,X)and with a monotonized point s∈S.Suppose a set-valued mapping
F:S→S\ {∅}satisfies the following conditions:
. Fis ordered metric increasing upward onSwith respect to the points∈S; . For everys∈S,the set{dX(u,s) :u∈F(s)}is a totally inductive subset ofX;
. The sets∈S{dX(u,s) :u∈F(s)}is a chain-complete subset ofX;
. There is an elements∈Ssuch thatdX(s,s)XdX(u,s),for someu∈F(s).
Then F has a fixed point.
Proof We define the ordered metric image set for the given mappingFas follows:
A(F) =dX
s,s:s∈S, there isu∈F(s) such thatdX
s,sXdX
u,s.
From condition ,dX(s,s)∈A(F); and therefore,A(F) is a nonempty subset ofX. For anydX(s,s)∈A(F), ifu∈F(s) satisfyingdX(s,s)XdX(u,s), then from condition andu∈F(s), there isv∈F(u) such thatdX(u,s)XdX(v,s). It impliesdX(u,s)∈A(F). So we get
dX
s,s∈A(F) and u∈F(s) withdX
s,sXdX
u,s
⇒ dX
u,s∈A(F). ()
Next, we show thatA(F) is inductive.
For any given chain{xα} ⊂A(F), suppose thatxα=dX(sα,s) withuα∈F(sα) such that
dX
sα,s
Xd
X
uα,s
, for allα. ()
From condition ,{dX(sα,s)}exists and is in
s∈S{dX(u,s) :u∈F(s)}. So there ares∈S andu∈F(s) such that
dX
sα,s
=dX
It implies
dX
sα,s
Xd
X
u,s, for allα. ()
For everyα, sinceuα∈F(sα), from condition , () implies that there isvα∈F(u) such
that
dX
uα,s
Xd
X
vα,s
.
From (), we have
dX
sα,s
Xd
X
vα,s
withvα∈F(u), for everyα. ()
From condition ,{dX(v,s) :v∈F(u)}is a totally inductive subset ofX. Then () implies that the chain{dX(sα,s)}has an upper bound in{dX(v,s) :v∈F(u)}, which is denoted by
dX(v,s) withv∈F(u). Hence
dX
sα,s
Xd
X
v,s, v∈F(u), for allα. ()
From (), it implies
dX(u,s) =
dX
sα,s
Xd
X
v,s withv∈F(u).
It follows that{dX(sα,s)}=dX(u,s)∈A(F). It implies that the chain{dX(sα,s)}has an
upper bound inA(F); and therefore,A(F) is inductive.
Applying Zorn’s lemma,A(F) has a maximal element, saydX(x∗,s), for some s∗∈S, which satisfies
dX
s∗,sXdX
u∗,s, for someu∗∈Fs∗. ()
From (), we havedX(u∗,s)∈A(F). SincedX(s∗,s) is a maximal element ofA(F), from (), it implies that
dX
u∗,s=dX
s∗,s. ()
Since (S,dX) is monodromy with respect to this elements∈S, () implies thats∗=u∗∈
F(s∗). Hences∗is a fixed point ofF.
Recall that every Riesz space (vector lattice) (X,X) can be considered as an ordered metric space with the ordered metric induced by its ordered absolute values. The positive cone of (X,X) is denoted by
X+=x∈X:xX.
with ∈Sas a monotonized point. It satisfies, for anys∈S,dX(s, ) =s; and therefore, for anys,s∈S,dX(s, )XdX(s, ) if and only ifsXs. Hence, if a set-valued mapping
F:S→S\ {∅}is ordered metric increasing upward, with respect to point , then it is ordered increasing upward, that is, whenever for any givens,s∈S,sXsimplies that, for everyt∈F(s), there ist∈F(s) satisfyingtXt. As a consequence of Theorem ., we have the following.
Corollary . Let(S,dX)be a nonempty subset of the positive cone of a Riesz space(X,X),
which contains .Suppose a set-valued mapping F:S→S\ {∅}satisfies the following conditions:
. Fis ordered increasing upward onS;
. for everys∈S,the setF(s)is a totally inductive subset ofX; . the sets∈SF(s)is a chain-complete subset ofX;
. there is an elements∈Ssuch thatsXu,for someu∈F(s).
Then F has a fixed point.
Remark . From Lemma ., Theorem . and Corollary . still hold if condition
is, respectively, replaced by
For everys∈S, the set{dX(u,s) :u∈F(s)}, orF(s), is an inductive subset ofXwith finite
number of maximal elements.
Considering single-valued mappings as special cases of set-valued mappings, the fol-lowing result follows immediately from Theorem ..
Corollary . Let(S,dX)be a monodromy ordered metric space with respect to an ordered
vector space(X,X)and with a monotonized point s∈S.If a single-valued self-mapping
f on S satisfies the following conditions:
. f is ordered metric increasing onSwith respect to the points∈S; . The set{dX(f(s),s) :s∈S}is a chain-complete subset ofX;
. There is an elements∈Ssuch thatdX(s,s)XdX(f(s),s).
Then f has a fixed point.
5 Applications to best ordered approximation problems
Since Fan [] proved a best approximation theorem by applying fixed point theorem on normed linear spaces, many authors have extended this theorem to more general topo-logical spaces and have provided applications to approximation theory (see [–]). As applications of the fixed point theorems proved in previous section, in this section, we ex-tend the concept of best approximation from metric spaces to ordered metric spaces and give some best ordered approximation theorems.
Definition . Let (S,dX) be an ordered metric space with respect to an ordered vector
space (X,X) andKa nonempty subset ofS. Letf:K→Sbe a map. An elementu∈Kis called a best ordered approximation off onK, if it satisfies
dX
u,f(u)=mindX
wheremin{dX(s,f(u)) :s∈K}is the smallest (minimum) element of the set{dX(s,f(u)) :
s∈K}with respect to the orderingXonX.
For any given single-valued mapf :K→S, define a set-valued mapTf:K→Kby
Tf(t) =
v∈K:dX
v,f(t)=mindX
s,f(t):s∈K, for allt∈K.
This mappingTf is called the approximating mapping of the mapf.
It is worthy to note that if (S,dX) is monodromy, then, for any given mapf :K→Sand fort∈K, the value of its approximating mappingTf(t) is either∅or a singleton.
As a consequence of Corollary ., we have the following.
Theorem . Let(S,dX)be a monodromy ordered metric space with respect to an ordered
vector space(X,X)and with a monotonized point s∈S and K a nonempty order-metric
complete subset of S containing s.Let f :K →S be a map.Suppose the approximating mapping Tf of f satisfies the following conditions:
. Tf(s)=∅;and therefore it is a singleton,for everys∈K;
. Tf is ordered metric increasing onKwith respect to the points∈K;
. the set{dX(Tf(s)) :s∈K}is a chain-complete subset ofX;
. there is an elements∈Ksuch thatdX(s,s)XdX(f(s),s).
Then f has a best ordered approximation on K.
Proof Sinces∈K, it is easily to see that (K,dX) is also a monodromy ordered metric space with respect to an ordered vector space (X,X) and with this monotonized pointss∈S. Condition implies that the mapTf :K→Sis well defined. From conditions - in this theorem, the mapTf satisfies all conditions in Corollary .. Then it follows thatTf has a fixed pointss∗=Tf(s∗). It implies
dX
s∗,fs∗=mindX
s∗,fs∗:s∈K.
Hences∗is a best ordered approximation off onK.
Theorem . Let(S,dX)be a monodromy ordered metric space,with respect to a
general-ized Archimedean ordered vector space(X,X)and with a monotonized point s∈S and K a nonempty order-metric complete subset of S containing s.Let f :K→S be a map.
Suppose the approximating mapping Tf:K→Kof f satisfies the following conditions:
. Tf(s)=∅,for everys∈K;
. there is a positive numberα< ,such thatdX(Tf(s),Tf(s))XαdX(s,s),for all
s,s∈S.
Then f has a best ordered approximation on K.
Definition . Let (S,dX) be an ordered metric space with respect to an ordered vector space (X,X) andKa nonempty subset ofS. Letf:K→Sbe a map. An elementu∈Kis called an extended best ordered approximation off onK, if it satisfies
dX
u,f(u)∈MindX
s,f(u):s∈K,
whereMin{dX(s,f(u)) :s∈K}is the set of minimal elements of the set{dX(s,f(u)) :s∈K} with respect to the orderingXonX.
It is clear that even in the case if (S,dX) is a monodromy, for any given mapf :K→S and fort∈K, the value of its approximating mappingTf(t) may not be a singleton. As a consequence of Theorem ., we have the following.
Theorem . Let(S,dX)be a monodromy ordered metric space with respect to an ordered
vector space(X,X)and with a monotonized point s∈S and K a nonempty subset of S
containing s.Let f :K→S be a map.Suppose the approximating mapping Tf :K→Kof
f satisfies the following conditions:
. Tf(s)=∅,for everys∈K;
. Tf is ordered metric increasing upward onKwith respect to the points∈K;
. for everys∈K,the set{dX(u,s) :u∈Tf(s)}is a totally inductive subset ofX;
. the sets∈K{dX(u,s) :u∈Tf(s)}is a chain-complete subset ofX;
. there is an elements∈Ksuch thatdX(s,s)XdX(u,s),for someu∈Tf(s).
Then f has a best ordered approximation on K.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors are equal contributors.
Author details
1Department of Mathematics, Shawnee State University, Second Street, Portsmouth, Ohio, USA.2Department of Applied
Mathematics, Nanjing University of Economic and Finance, Wenyuan Road, Nanjing, 210046, China.3Department of
Applied Mathematics, Nanjing University of Science and Technology, Xiaolingwei Street, Nanjing, Jiangsu 210094, China.
Acknowledgements
The authors are grateful for those who have offered their valuable suggestions on improving our work.
Received: 8 January 2014 Accepted: 21 April 2014 Published:06 May 2014 References
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Cite this article as:Li et al.:Fixed point theorems on ordered vector spaces.Fixed Point Theory and Applications
