R E S E A R C H
Open Access
Fixed points of monotone mappings and
application to integral equations
Mostafa Bachar
1*and Mohamed Amine Khamsi
2,3*Correspondence: [email protected]
1Department of Mathematics, King Saud University, Riyadh, Saudi Arabia
Full list of author information is available at the end of the article
Abstract
In this work, we discuss the existence of fixed points of monotone nonexpansive mappings defined on partially ordered Banach spaces. This work is a continuity of the previous works of Ran and Reurings, Nietoet al., and Jachimsky done for contraction mappings. As an application, we discuss the existence of solutions to an integral equations.
MSC: Primary 46B20; 45D05; secondary 47E10; 34A12
Keywords: fixed point; integral equation; Krasnoselskii iteration; Lebesgue measure; monotone mapping; nonexpansive mapping
1 Introduction
Banach’s contraction principle [] is remarkable in its simplicity, yet it is perhaps the most widely applied fixed point theorem in all of analysis. This is because the contractive condi-tion on the mapping is simple and easy to test, because it requires only a complete metric space for its setting, and because it finds almost canonical applications in the theory of dif-ferential and integral equations. Over the years, many mathematicians tried successfully to extend this fundamental theorem. Recently a version of this theorem has been given in partially ordered metric spaces [, ] (see also [, ]) and in metric spaces with a graph []. In this work, we discuss the case of nonexpansive mappings defined in partially ordered Banach spaces. Nonexpansive mappings are those which have Lipschitz constant equal to . The fixed point theory for such mappings is rich and varied. It finds many applications in nonlinear functional analysis []. It is worth mentioning that such investigation is new and has never been carried.
2 Monotone nonexpansive mappings
Let (X, · ) be a Banach vector space. Assume that we have a partial orderdefined onX such that order intervals are convex andτ-closed, whereτ is a Hausdorff topology onX. Recall that an order interval is any of the subsets [a,b] ={x∈X;axb}, [a,→) ={x∈ X;ax}, (←,a] ={x∈X;xa}for anya,b∈X.
Definition . LetCbe a nonempty subset ofX. LetT:C→Cbe a map. () Tis said to be monotone ifT(x)T(y)wheneverxyfor anyx,y∈C. () Tis said to be monotone nonexpansive if and only ifT is monotone and
T(x) –T(y)≤ x–y, wheneverxy.
The pointx∈Cis called a fixed point ofTifT(x) =x. The set of fixed points ofT will be denoted byFix(T).
Throughout the paper we assume thatC is convex and bounded not reduced to one point. LetT:C→Cbe a monotone nonexpansive mapping. Fixλ∈(, ) andx∈C. The Krasnoselskii [, ] iteration sequence{xn} ⊂Cis defined by
xn+=λxn+ ( –λ)T(xn), n≥. (KIS)
The following lemma holds.
Lemma . Under the above assumptions,if we assume that xT(x),then we have
xnxn+T(xn)T(xn+) (KI)
for any n≥.Moreover,if{xn}has two subsequences whichτ-converge to z and w respec-tively,then we must have z=w.
Proof First note that ifxyholds, then we havexλx+ ( –λ)yyfor anyx,y∈X since order intervals are convex. Therefore it is enough to only provexnT(xn) for any n≥. By assumption, we havexT(x). Assume thatxnT(xn) forn≥. Then we havexnλxn+ ( –λ)T(xn)T(xn),i.e.,xnxn+T(xn). SinceTis monotone, we get T(xn)T(xn+). By induction, we conclude that the inequalities (KI) hold for anyn≥. Next let {xφ(n)} be a subsequence of{xn}which τ-converges toz. Clearly,{[xn,→);n∈ N}is a decreasing family of sets. Consequently, ifUz ∈τ is a neighborhood ofz, then Uz∩[xn,→)=∅for anyn∈N. Therefore,zbelongs to all sets [xn,→) as they are closed. Letwbe theτ-limit of another subsequence of{xn}. IfUw∈τ is a neighborhood ofw, thenUwcontains many points from the sequence{xn}sincewis aτ-limit of one of its subsequences. HenceUw∩(←,z]= ∅. Therefore,wbelongs to (←,z] as it is closed,i.e., wz. By reversing the roles ofzandw, we getzw. The properties of the partial order
will forcez=was claimed.
Remark . Note that under the assumptions of Lemma ., if we assumeT(x)x, then we will have
T(xn+)T(xn)xn+xn
for anyn≥. The conclusion on theτ-convergence limits of{xn}will also hold. The following result is found in [, ].
Proposition . Under the above assumptions,we have
+n( –λ)T(xi) –xi
for any i,n∈N.This inequality implies
lim
n→+∞xn–T(xn)= .
Proof The first part of this proposition is easy to prove via an induction argument on the indexi. As for the second part, note that{xn–T(xn)}is decreasing. Indeed we have xn+–xn= ( –λ)(T(xn) –xn) for anyn≥. Therefore{xn–T(xn)}is decreasing if and only if{xn+–xn}is decreasing, which holds since
xn+–xn+ ≤λxn+–xn+ ( –λ)T(xn+) –T(xn)≤ xn+–xn
for anyn≥. Setlimn→+∞xn–T(xn)=R. Then we leti→+∞in the inequality (GK) to obtain
+n( –λ)R≤δ(C)
for anyn∈N, whereδ(C) =sup{x–y,x,y∈C}< +∞. Hence
R≤ δ(C)
( +n( –λ)), n= , , . . . ,
which impliesR= ,i.e.,limn→+∞xn–T(xn)= .
Before we state the main result of this work, let us recall the definition of Opial condi-tion [].
Definition . X is said to satisfy theτ-Opial condition if for any sequence{yn}inX whichτ-converges toy, we have
lim sup
n→+∞
yn–y<lim sup n→+∞
yn–z
for anyz∈Xsuch thatz=y.
Now we are ready to state the main result of this section.
Theorem . Let X be a Banach space.Letτ be a topology on X such that X satisfies the
τ-Opial condition.Letbe a partial order on X such that order intervals are convex and
τ-closed.Let C be a bounded convexτ-compact nonempty subset of X.Let T:C→C be a monotone nonexpansive mapping.Assume that there exists x∈C such that xand T(x) are comparable.Then T has a fixed point.
Proof Without loss of any generality, we assume thatxT(x). Consider the (KIS) se-quence{xn}which starts atx. SinceCisτ-compact, then{xn}will have a subsequence
{xkn}whichτ-converges to some pointw∈C. Lemma . implies that{xn}τ-converges to wandxnwfor anyn∈N. Consider the type function
r(x) =lim sup
n→+∞
Then Proposition . implies r(x) =lim supn→+∞T(xn) –xfor any x∈C. Since T is monotone nonexpansive, we get
rT(w)=lim sup
n→+∞
T(xn) –T(w)≤lim sup
n→+∞
xn–w=r(w).
In fact we haver(T(x))≤r(x) for anyx∈Csuch thatxnandxare comparable for any n∈N. Finally, ifXsatisfies theτ-Opial condition, then we must haveT(w) =w,i.e.,wis a
fixed point ofT.
The following results are direct consequences of Theorem ..
Corollary . Let C be a bounded closed convex nonempty subset of lp, <p< +∞.Let
τ be the weak topology.Consider the pointwise partial ordering in lp,i.e., (αn)(βn)iff
αn≤βnfor all n≥.Then any monotone nonexpansive mapping T:C→C has a fixed point provided there exists a point x∈C such that xand T(x)are comparable.
Remark . The case ofp= is not interesting for the weak topology sincelis a Schur Banach space. But if we consider the weak* topologyσ(l,c) onlor the pointwise conver-gence topology, thenlsatisfies the Opial condition for these topologies. Note that these two topologies are Hausdorff. In this case we have a similar conclusion of Corollary . forl.
Recall the definitions oflpandcspaces: (i) lp={(αn)∈RN,
n|αn|p< +∞}for≤p< +∞; (ii) c={(αn)∈RN,limn→+∞αn= }.
3 Application to integral equations
Let us consider the following integral equation of the form
x(t) =g(t) +
Ft,s,x(s)ds, t∈[, ], (IE)
where
(i) gis inL([, ],R),
(ii) F: [, ]×[, ]×L([, ],R)→Ris measurable and satisfies the condition
≤F(t,s,x) –F(t,s,y)≤x–y, (.)
wheret,s∈[, ], andx,y∈L([, ],R)such thaty≤x. Recall that for anyu,v∈L([, ],R), we have
u≤v ⇐⇒ u(t)≤v(t) almost everywheret∈[, ].
Assume that there exists a non-negative functionh(·,·)∈L([, ]×[, ]) andM< such that
F(t,s,x)≤h(t,s) +M|x|, (.)
wheret,s∈[, ] andx∈L([, ],R). Let
B=y∈L[, ],R, such thatxL([,],R)≤ρ ,
whereρis sufficiently large,i.e.,Bis the closed ball ofL([, ],R) centered at with ra-diusρ. Consider the operator defined by
F(t)(y)(s) =Ft,s,y(s), (.)
and define the operatorJ:L([, ],R)→L([, ],R) by
(Jy)(t) =g(t) +
F(t)(y)(s)ds. (.)
We haveJ(B)⊂B. Indeed letx∈B, then by using the Cauchy-Schwarz inequality, condi-tion (.) and the quadratic inequality (a+b)≤a+ bfor anya,b∈R, we have
JxL([,],R) =
Jx(t)dt
=
g(t) +
F(t)(x)(s)ds
dt
≤
g(t)dt+
F(t)(x)(s)ds dt
≤
g(t)
dt+
h(t,s) +Mx(s) ds dt
≤
g(t)dt+
h(t,s)ds dt+ M
x(s)ds dt
=
g(t)dt+
h(t,s)ds dt+ M
x(s)ds
=
g(t)dt+
h(t,s)ds dt+ MxL([,],R)
≤
g(t)dt+
h(t,s)ds dt+ Mρ.
SinceM< /, chooseρsuch that
( – M)
g(t)dt+ ( – M)
h(t,s)ds dt≤ρ,
Cauchy-Schwarz inequality, we have
Jx–JyL([,],R) =
Jx(t) –Jy(t)dt
=
F(t)(x)(s) –F(t)(y)(s)ds
dt
≤
x(s) –y(s)ds
dt
≤
x(s) –y(s)ds=x–yL([,],R),
which implies that J is a monotone nonexpansive operator as claimed. In order to use Theorem ., we need to check its assumptions. First note thatX=L([, ],R) is a Hilbert space. If we chooseτ to be the weak topology, thenXsatisfies the weak Opial condition. It is easy to check that order intervals are convex. In order to show that order intervals are closed, we will show that if{un}is a non-negative sequence of elements inXwhich converges weakly tou, thenuis positive. Leta< , then the setA={t∈[, ];u(t)≤a} has measure . Indeed, we have
A
u(t)dt= lim
n→+∞
A un(t)dt
because of weak convergence. So
≤ lim
n→+∞
A
un(t)dt=
A
u(t)dt≤am(A)≤.
Hencem(A) = . Set
D= n≥
t∈[, ];u(t)≤– n
.
ThenDhas measure , which implies thatu(t)≥ for almost everyt∈[, ]. Using The-orem ., we get the following result.
Theorem . Under the above assumptions,we conclude that
(i) the integral equation(IE)has a non-negative solution provided we assume that
g(t) +F(t,s, )ds≥for almost everyt∈[, ](which impliesJ()≥); (ii) the integral equation(IE)has a non-positive solution provided we assume that
g(t) +F(t,s, )ds≤for almost everyt∈[, ](which impliesJ()≤).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Author details
Acknowledgements
The authors would like to extend their sincere appreciation to the Deanship of Scientific Research at King Saud University for funding this Research group No. (RG-1435-079).
Received: 27 February 2015 Accepted: 23 June 2015 References
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