R E S E A R C H
Open Access
Fixed points of multi-valued monotone
operators and the solvability of a fractional
integral inclusion
Yuqiang Feng
1,2*and Yuanyuan Wang
1,2*Correspondence:
yqfeng6@126.com
1School of Science, Wuhan
University of Science and Technology, Wuhan, 430065, P.R. China
2Hubei Province Key Laboratory of
Systems Science in Metallurgical Process, Wuhan, 430065, P.R. China
Abstract
Based on the characterizations of reproducing cones, some fixed point theorems for multi-valued increasing, decreasing, and mixed monotone operators are established. As an application, the solvability of a fractional integral inclusion is discussed.
Keywords: reproducing cone; multi-valued operator; fixed point; fractional integral inclusion; monotone
1 Introduction
Single-valued monotone operators have been widely investigated. The results on the exis-tence of fixed points for single-valued monotone operators are bounteous and successful and have found various applications to nonlinear integral equations and differential equa-tions. For details, we can refer to [–] and the references therein.
It is well known that mixed monotone operators were introduced by Guo and Laksh-mikantham [] in . Later, Bhaskar and LakshLaksh-mikantham [] introduced the notation of coupled fixed point and proved some coupled fixed point results under certain condi-tions, in a complete metric space endowed with a partial order. Their study not only has important theoretical meaning but also wide applications in engineering, nuclear physics, biological chemistry technology, economics, etc. (see [–] and the references therein).
Very recently, Harjani, Lopez and Sadarangani [] have established the existence results of coupled fixed point for mixed monotone operators and further obtained their applica-tions to integral equaapplica-tions. Then in [], Bu, Feng, and Li extended the study to mixed monotone ternary operators.
It is natural to extend these studies to the multi-valued case. In , Nishniannidze [] introduced the multi-valued monotone operators. Nguyen and Nguyen [] investi-gated the fixed points for multi-valued increasing operators and then in [], a fixed point theorem for a multi-valued increasing operator was established and applied to a discon-tinuous elliptic equation. Huang and Fang [] extended the mixed monotone operators to the multi-valued case, the obtained result was applied to a class of integral inclusions. For some recent fixed point theorems for a multi-valued monotone operator, a mixed mono-tone operator, and their applications in differential equations and differential inclusions,
we can refer to [–] and the references therein for details. All these papers need the existence of a lower or an upper solution to the operator inclusion.
Motivated by the above work, in this paper, we shall use the properties of reproduc-ing cones to establish some fixed point theorems for multi-valued monotone and mixed monotone operators. Compared with the previous work, we remove the requirement of the existence of a lower or upper solution. When the operator is single-valued and mixed monotone, we get an existence and uniqueness result.
This paper is organized as follows: in Section , some basic knowledge and the proper-ties of a reproducing cone are presented; then in Section , the existence of a fixed point for multi-valued monotone operators is established; in Section , coupled fixed point the-orems are presented for single-valued and multi-valued mixed monotone operators. In the final section, as applications of our results, the solvability of a fractional integral inclusion is discussed.
2 Preliminaries
In this section, we recall some standard definitions and notations needed in the following section. For convenience of the reader, we suggest that one refers to [, ] for details.
At the beginning of this section, let us recall some concepts of the theory of cones in Banach spaces. These concepts play an important role in the remainder of this paper.
LetXbe a Banach space, a closed convex setP⊂Xis called acone, ifx∈Pandx= impliesαx∈Pforα≥ andαx∈/Pforα< . A cone defines apartial orderin the Banach spaceX: we writex≤yory≥xify–x∈P. The relation enjoys the following properties: inequalities may be multiplied by a nonnegative numbers; inequalities of the same kind may be added by terms; one may pass to the limit in inequalities;x≤yandy≤ximplies x=y. Denote –P={–x|x∈P}, then –Pis a cone too.
It is well known that ifXbe a partially ordered Banach space endowed with partial or-der≤, then the subsetP={x∈X|≤x}is a cone.
Definition .[] A coneP is callednormal, if there is aK> , such that ≤x≤y impliesx ≤KyandKdoes not depend on xandy. Any suchK=K(P) is called a normal constantofP. A conePis calledreproducing, if eachx∈Xadmits a presentation x=u–v(u,v∈P). The elementsu,vare, of course, not unique.
For the details of cone theory, see [] and references therein.
Lemma .[] Let X be a Banach space,P⊂X be a cone.The following assertions are equivalent:
() Pis reproducing;
() every pairx,y∈Xhas a lower bound; () every pairx,y∈Xhas an upper bound; () ∀x∈X,there existsu≥such thatx≤u; () ∀x∈X,there existsu≤such thatx≥u.
Definition .[] LetXbe a topology space, ‘≤’ be a partial order endowed onX, letA, Bbe two nonempty subsets ofX, the relations betweenAandBare defined as follows:
Definition . Let (X,≤) be an ordered Banach space, the partial order ‘≤’ is induced by coneP.
() A multi-valued operatorT:X→X\ {φ}is called increasing, if for∀x,y∈X,x≤y
impliesT(x)≺T(y).
() A multi-valued operatorT:X→X\ {φ}is called decreasing, if for∀x,y∈X,x≤y
impliesT(y)≺T(x).
() A multi-valued operatorT:X×X→X\ {φ}is called mixed monotone, if for ∀x,x,y,y∈X,x≤x,y≤yimpliesT(x,y)≺T(x,y).
Definition .
() LetT:X→X\ {φ}be a multi-valued operator,x∈Xis called a fixed point ofT, if
x∈T(x).
() AssumeT:X×X→X\ {φ}is a multi-valued operator,(x,y)∈X×Xis called a
coupled fixed point ofT, ifx∈T(x,y)andy∈T(y,x).
3 Fixed points for multi-valued monotone operators
In this section, some fixed point theorems for multi-valued increasing and decreasing op-erators are proved in partial ordered Banach space.
Throughout this paper, we assume thatXis a Banach space,Pis a normal and repro-ducing cone inXand the partial order ‘≤’ is induced by the coneP.
Theorem . Suppose the multi-valued operator T:X→X\ {φ}satisfies the following conditions:
() For anyx∈X,T(x)is a nonempty and closed subset ofX.
() There exists a linear operatorL:X→Xwith spectral radiusr(L) < ,L(P)⊂Psuch that,for anyx,y∈X,x≤yimplies:
(i) for anyu∈T(x),there existsv∈T(y)satisfying
≤v–u≤L(y–x),
(ii) for anyv∈T(y),there existsu∈T(x)satisfying
≤v–u≤L(y–x).
Then T has a fixed point in X.
Proof The proof is given in three steps.
Step . There existsx∈X, such that{x} ≺T(x). In fact,
() if{} ≺T(), thenx= ; () if{} ≺T()does not hold.
Lety∈T(). SincePis reproducing, by Lemma ., there isv∈(–P), such thatv≤y.
Due to the assumptionr(L) < ,L(P)⊂P, we know that the equation
(I–L)x= –v
Letx= –z, by assumption ()(ii), fory∈T(), there existsu∈T(x)such that
≤y–u≤L( –x),
hence,
u≥y–L(z).
Noting thatv≤yand(I–L)(z) = –v, we have
u≥y–L(z)≥v–L(z) = –z=x,
which implies{x} ≺T(x).
Step . There exists an increasing and fundamental sequence{xn} ⊂X.
In fact, letx=u, due to assumption ()(i), there existsx∈T(x)such that
≤x–x≤L(x–x).
Repeating the arguments above for the pairx,xin place ofx,xand so on, we can find an increasing sequence{xn}satisfying
xn∈T(xn–), ≤xn+–xn≤Ln(x–x).
Sincelimn→∞(Ln)
n =r(L)andr(L) < , we haveLn ≤qnfor some
q∈(, )and for all sufficiently largen. At this moment, we get
xn–xn– ≤KLn· x–x ≤Kqnx–x,
whereKis the normal constant of coneP.
This implies{xn}is fundamental. SinceXis complete, there exists a unique x∗∈Xsuch thatxn→x∗.
Step . x∗∈T(x∗).
Sincexn–≤x∗,xn∈T(xn–), by assumption (i), there existsyn∈T(x∗), such
that
≤yn–xn≤L
x∗–xn–
.
Due to the fact thatPis normal, we have
yn–xn ≤KL ·x∗–xn–,
which implies
lim
n→∞yn=nlim→∞(yn–xn) +nlim→∞
xn–x∗
+x∗=x∗.
Noting thatT(x∗)is closed, we knowx∗∈T(x∗).
Remark .
. In Theorem ., the assumptions ()(i) and ()(ii) imply thatT(x)≺T(y)forx≤y, i.e.Tis a multi-valued increasing operator.
. It should be noticed that Theorem . cannot ensure the uniqueness of a fixed point. For example, letX=R,x=|x|,P={x∈R|x≥}, thenXis a Banach space andPa normal and reproducing cone.
DefineT:X→X\ {φ},L:X→Xas follows:
T(x) =
u∈X x–
≤u≤
x+
, x∈X,
L(x) = x.
All conditions of Theorem . are satisfied. It is easy to verify that the fixed point set ofTis the interval[–, ].
. In the special case whenTis single-valued, assumption () of Theorem . is naturally satisfied. Assumption () is simplified thus: there exists a linear operator
L:X→Xwith spectral radiusr(L) < ,L(P)⊂Psuch that
T(y) –T(x)≤L(y–x), ∀x≤y.
In this case, for the fixed point ofT we have existence and uniqueness; see [].
Remark . In Theorem ., assumptions (), () are sufficient, without one of them, the existence of a fixed point cannot be ensured. For example, letX=R,x=|x|,P={x∈R| x≥}, thenXis a Banach space andPa normal and reproducing cone.
. DefineT:X→X\ {φ},L:X→Xas follows:
T(x) =
⎧ ⎨ ⎩
(–, ), x≤–orx≥,
(–,x)∪(x, ), – <x< ,
L(x) = x.
Then assumption ()(i), ()(ii) of Theorem . are satisfied. ButT(x)is not closed. It is obvious thatThas no fixed point.
. DefineT:X→X\ {φ}as T(x) = [x+ ,∞).
ThenTsatisfies assumptions () and ()(ii), whereL(x) =λxfor anyλ∈(, ). But
T cannot satisfy assumption ()(i). It is obvious thatThas no fixed point. . DefineT:X→X\ {φ}as
T(x) = (–∞,x– ].
ThenTsatisfies assumptions () and ()(i), whereL(x) =λxfor anyλ∈(, ). But
In the following, we give a fixed point theorem for multi-valued decreasing operator.
Theorem . Suppose the multi-valued operator T:X→X\ {φ}satisfies the following conditions:
() For anyx∈X,T(x)is a nonempty and closed subset ofX.
() There exists a positive constantc∈(, )such that,for anyx,y∈X,x≤yimplies: (i) for anyu∈T(x),there existsv∈T(y)satisfying
–c(y–x)≤v–u≤,
(ii) for anyv∈T(y),there existsu∈T(x)satisfying
–c(y–x)≤v–u≤.
Then T has a fixed point in X.
Proof Define a new mappingS:X→X\ {φ}as follows:
S(x) =
w∈Xw= c +cx+
+cu,u∈T(x)
,
i.e.
S(x) = c +cx+
+cT(x).
Then it is easy to verify thatSsatisfies:
() For anyx∈X,S(x)is a nonempty and closed subset ofX. () For anyx,y∈X,x≤yimplies:
(i) for anyu∈S(x), there existsv∈S(y)satisfying
≤v–u≤ c
+c(y–x),
(ii) for anyv∈S(y), there existsu∈S(x)satisfying
≤v–u≤ c
+c(y–x).
By Theorem ., there existsx∗∈Xsuch that
x∗∈Sx∗
i.e.
x∗∈ c +cx
∗+ +cT
x∗,
which implies
x∗∈Tx∗.
4 Coupled fixed points for mixed monotone operators
In this section, we shall establish the existence and uniqueness of coupled fixed point for single-valued mixed monotone operators, and then we extend the study to the multi-valued case.
To verify the main results, we need the following lemma.
Lemma . If(X, · )is a Banach space,P is a normal and reproducing cone in X,then
() X×Xis a Banach space with the norm(x,y)=max{x,y}. () P×(–P)is a normal and reproducing cone inX×X.
Proof We only prove assertion (). Noting that a partial orderofX×Xcan be induced byP×(–P), we have
(x,y)(x,y) ⇐⇒ x–x∈P,y–y∈(–P) ⇐⇒ x≤x,y≤y.
(i) If(, )(x,y)(x,y), then≤x≤x,≥y≥y, or equivalently,
≤–y≤–y. SincePis normal,
x ≤Kx, y ≤Ky,
whereKis the normal constant ofP. Hence, we know
(x,y)=max
x,y
≤Kmaxx,y
=K(x,y),
which impliesP×(–P)is normal with normal constantK.
(ii) For arbitrary(x,y)∈X×X, sincePis reproducing, there existu,v,u,v∈P, such thatx=u–v,y=u–v, or equivalently,y= (–v) – (–v). Then
(x,y) = (u, –v) – (u, –v)∈P×(–P) –P×(–P),
which impliesP×(–P)is a reproducing cone inX×X.
Theorem . Suppose T:X×X→X is a mixed monotone operator.Assume there exist two linear operators L,S:X→X withL+S< ,L(P)⊂P,S(P)⊂P such that,for any x,x,y,y∈X,x≤x,y≤y
T(x,y) –T(x,y)≤L(x–x) +S(y–y).
Then T has a unique coupled fixed point(x,y)in X×X.Moreover,for every(x,y)∈X×X,
lim
n→∞T
n(x,y) =x, lim n→∞T
n(y,x) =y.
Proof Define a new mappingF:X×X→X×Xas follows, for (x,y)∈X×X:
ThenT has a coupled fixed point if and only ifFhas a fixed point. WhenTis mixed monotone, then∀x,x,y,y∈X,
x≤x,y≤y ⇒ T(x,y)≤T(x,y),T(y,x)≥T(y,x),
which means
(x,y)(x,y) ⇒ F(x,y)F(x,y)
i.e. Fis increasing.
Furthermore, by assumption, for anyx,x,y,y∈X,x≤x,y≤y
T(x,y) –T(x,y)≤L(x–x) +S(y–y) =L(x–x) –S(y–y),
then
T(y,x) –T(y,x)≤L(y–y) +S(x–x),
or equivalently,
T(y,x) –T(y,x)≥–L(y–y) –S(x–x) = –S(x–x) +L(y–y),
which implies
F(x,y) –F(x,y)
(x,y) – (x,y)
, ∀(x,y)(x,y),
where:X×X→X×Xis defined as
(x,y) =L(x) –S(y), –S(x) +L(y).
Note that
() is a linear operator and(P×(–P))⊂P×(–P).
In fact, sinceL,Sis linear,is a linear operator onX×X.
If(x,y)∈P×(–P),i.e.x≥,y≤, due to the positivity ofL,S, we haveL(x)≥,
L(y)≤,S(x)≥,S(y)≤, then
L(x) –S(y)≥, –S(x) +L(y)≤
i.e.
(x,y)∈P×(–P).
In fact,
= sup
(x,y)∈X×X,(x,y)=
(x,y)
= sup
(x,y)∈X×X,(x,y)=
maxL(x) –S(y),L(y) –S(x)
≤ sup
(x,y)∈X×X,(x,y)=
maxL(x)+S(y),L(y)+S(x)
≤ sup
(x,y)∈X×X,max{x,y}=
maxLx+Sy,Ly+Sx
≤ L+S.
Then, by [] Theorem .,Fhas a unique fixed point (x,y)∈X×X. Moreover, for every (x,y)∈X×X,
lim
n→∞F
n(x,y) = (x,y).
Hence,T has a unique coupled fixed point (x,y) inX×X. Moreover, for every (x,y)∈ X×X,
lim
n→∞T
n(x,y) =x, lim n→∞T
n(y,x) =y.
This ends the proof.
Remark . If, in addition, the positive linear operatorsL,Sare commutative,i.e. LS=SL, then the requirementL+S< can be relaxed tor(L) +r(S) < .
In this case, by mathematical induction, we have
n(x,y) =L+Sn(x),L+Sn(y), ∀(x,y)∈X×X, then
n= sup
(x,y)∈X×X,(x,y)=
(x,y)
= sup
(x,y)∈X×X,(x,y)=
maxL+Sn(x),L+Sn(y)
≤L+Sn.
Hence, by Gelfand’s formula
r() = lim
n→∞() nn
= lim
n→∞()
nn
≤ lim
n→∞L
+Snn
=
≤r(L)+r(S)
≤r(L) +r(S)
< .
Then, by [] Theorem .,Fhas a unique fixed point (x,y)∈X×X.
Now, we extend the study to the multi-valued case.
In the same manner as the proof of Theorem . and Theorem ., we can verify the following coupled fixed point theorem for multi-valued mixed monotone operators.
Theorem . Suppose T:X×X→X\ {φ}is a multi-valued mixed monotone operator. Assume the following conditions are satisfied:
() For any(x,y)∈X×X,T(x,y)is a nonempty and closed subset ofX.
() There exist two linear operatorsL,S:X→X,L+S< ,L(P)⊂P,S(P)⊂Psuch that,for anyx,x,y,y∈X,x≤x,y≤yimply:
(i) for anyu∈T(x,y),there existsv∈T(x,y)
≤v–u≤L(x–x) +S(y–y),
(ii) for anyv∈T(x,y),there existsu∈T(x,y)
≤v–u≤L(x–x) +S(y–y).
Then T has a coupled fixed point in X×X.
5 Application to a fractional integral inclusion
Let (,) be a measurable space andXbe a Banach space. We use the notations
Pf(X) ={A⊂X|Ais nonempty and closed},
Pkc(X) ={A⊂X|Ais nonempty, compact and convex}.
A multi-valued mappingF:→Pf(X) is said to be measurable if, for everyx∈X,
ω→dx,F(ω)= inf
z∈F(ω)x–z
is measurable.
LetX=C[, ]. Forx∈X, letx=maxt∈[,]|x(t)|,P={x∈X|x(t)≥,t∈[, ]}, then
(X, · ) is a Banach space andPis a normal and reproducing cone inX.
Let <α< be a constant. Forx∈X, the Riemann-Liouville fractional integral operator Iαis defined as follows []:
Iαx(t) = (α)
t
Letf: [, ]×R→R\ {φ}be a multi-valued function. For eachu∈X, define the set of selections off by
Sf,u=
v∈Iα:v(s)∈fs,u(s)a.e.s∈[, ],
whereIαstands for the collection of all Riemann-Liouville fractional integral functions,
and further assume thatSf,uis nonempty for eachu∈x.
Now, let us consider the following fractional integral inclusion:
x(t)∈
t
(t–s)α–fs,x(s)ds,
where
t
(t–s)α–fs,x(s)ds=
t
(t–s)α–u(s)ds:u(s)∈Sf,x
.
Theorem . Assume the following conditions hold:
() f : [, ]×R→R\ {φ}satisfies the following hypotheses:
(i) f(t,x)inPkc(R)for(t,x)∈[, ]×R; (ii) for eachu∈X,t→f(t,u(t))is measurable;
(iii) for eacht∈[, ]andu∈C[, ],supx∈f(·,u(·))x ∈L.
() There exists a constantM> such that for anyx,y∈X,x≤yimplies: (i) for anyu∈t(t–s)α–f(s,x(s))ds∩C[, ],there exists
v∈t(t–s)α–f(s,y(s))ds∩C[, ]satisfying
≤v(t) –u(t)≤
t
(t–s)α–My(s) –x(s)ds,
(ii) for anyv∈t(t–s)α–f(s,y(s))ds∩C[, ],there exists
u∈t(t–s)α–f(s,x(s))ds∩C[, ]satisfying
≤v(t) –u(t)≤
t
(t–s)α–My(s) –x(s)ds.
Then the fractional integral inclusion has a solution in C[, ].
To prove the solvability of fractional integral inclusion, we need the following lemma.
Lemma . Iα is a linear, positive operator from X to X.Moreover,the spectral radius
r(Iα) = .
Proof The proof is given in two steps.
Step . IαmapsXintoX.
Whenh>
Iαu(t+h) –Iαu(t)= (α)
t+h
(t+h–τ)α–u(τ)dτ
– (α)
t
(t–τ)α–u(τ)dτ
≤ u (α)
t
(t+h–τ)α–– (t–τ)α–dτ
+ u
(α)
t+h(t+h–τ)α–dτ
= u (α)
|K|+
hα
α
.
ForK, we have the following estimation:
|K|=
t(t+h–τ)α–– (t–τ)α–dτ=hα
t
h
sα–– (s+ )α–ds.
() If≤t≤h,
|K|<hα
sα–– (s+ )α–ds<hαh α
α .
() Ift>h, then
|K|=
(α) t+h
(t+h–τ)α–u(τ)dτ– (α)
t
(t–τ)α–u(τ)dτ
=hα
sα–– (s+ )α–ds+hα
t
h
sα–– (s+ )α–ds
<h
α
α + ( –α)h
α
t
h
sα–ds
<h
α
α + ( –α)h
α
∞
sα–ds
=
+
α
hα.
By (), (), we have
lim
h→
Iαu(t+h) –Iαu(t)= .
In the same manner, we can provelimh→|Iαu(t+h) –Iαu(t)|= forh< . Hence,Iαu∈X. The linearity and positivity ofIαare obvious.
Step . r(Iα) = .
By the semi-group property of the Riemann-Liouville fractional integral operator, we have
Iαnu(t) =Inαu(t) = (nα)
t
hence
Iαn= sup
u=
max
≤t≤
(nα)t(t–τ)nα–u(τ)dτ
≤max
≤t≤
(nα) t
(t–τ)nα–dτ
=max
≤t≤
(nα)
nαt
nα
≤
(nα+ ).
Due to Stiring’s formula,
lim
n→∞
n
(nα+ ) = lim
n→∞
n
nα
e
nα√
nπ α +o()= lim
n→∞
nα
e
α
=∞.
By Gelfand’s formula,
rIα= lim
n→∞I
αnn≤ lim
n→∞
√n
(nα+ )
= .
Thenr(Iα) = .
This ends the proof.
Proof of Theorem. Define a multi-valued mappingT fromXto X\ {φ}as follows:
T(x)(t) =
t
(t–s)α–fs,x(s)ds, x∈X.
From assumption () we know thatT has nonempty values. Because of the Rädstrom embedding theorem [], it is easy to show
T(x)(t)∈Pkc(R), t∈[, ].
Then an application of Arzela and Ascoli’s theorem shows thatThas values inPkc(C[, ]).
Assumption () means there exists a linear operatorL:X→Xsuch that, for anyx,y∈X, x≤yimplies:
(i) for anyu∈T(x), there existsv∈T(y)satisfying
≤v–u≤L(y–x),
(ii) for anyv∈T(y), there existsu∈T(x)satisfying
≤v–u≤L(y–x),
whereL(x)(t) =Mt(t–s)α–x(s)ds. By Lemma .,r(L) = andL(P)⊂P. Due to Theo-rem .,T has a fixed point inx,i.e., the fractional differential inclusion has a solution.
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.
Acknowledgements
We are grateful to the anonymous referees for their helpful suggestions. This research is supported in part by the Doctoral Fund of Education Ministry of China (20134219120003) and the Key Program of Natural Science Foundation of China (71231007).
Received: 5 February 2016 Accepted: 18 May 2016 References
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