R E S E A R C H
Open Access
Positive fixed points for convex and
decreasing operators in probabilistic Banach
spaces with an application to a two-point
boundary value problem
Mohamed Jleli and Bessem Samet
**Correspondence: [email protected] Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh, 11451, Saudi Arabia
Abstract
We establish a fixed point theorem for decreasing and convex operators in a
probabilistic Banach space partially ordered by a normal cone. We give an application to the study of the existence and uniqueness of positive solutions to a two-point boundary value problem.
MSC: 47H07; 33E30
Keywords: positive fixed point; probabilistic Banach space; cone; convex and decreasing operator; boundary value problem
1 Introduction and preliminaries
The concept of probabilistic Banach spaces was introduced by Serstnev [] by adapting the idea of Menger [] to linear spaces. Fixed point theory in such spaces was studied and developed by many authors (see [–] and the references mentioned therein).
This paper deals with the existence and uniqueness of fixed points for a certain class of convex and decreasing operators defined in a probabilistic Banach space partially ordered by a cone.
For the sake of convenience, we first give some definitions and known results from the existing literature. For more details, we refer to [, ].
Definition . A functionf :R→Ris said to be a distribution function if it satisfies the following conditions:
(i) f is non-decreasing; (ii) f is left-continuous;
(iii) inft∈Rf(t) = andsupt∈Rf(t) = .
We denote byDthe set of all distribution functions.
Definition . A triangular norm, briefly a T-norm, is a mappingT: [, ]×[, ]→[, ] that is continuous and such that for everya,b,c,d∈[, ],
(i) T(a, ) =a; (ii) T(a,b) =T(b,a);
(iii) c≥a,d≥b⇒T(c,d)≥T(a,b); (iv) T(T(a,b),c) =T(a,T(b,c)).
As standard examples,Tm(a,b) =min{a,b}andTp(a,b) =abon [, ] are T-norms. Definition . LetXbe a real vector space,T be a T-norm andN:X→Dbe a given mapping. We say thatNis a probabilistic norm onXif the following conditions hold:
(i) Nx() = for everyx∈X; (ii) Nx(t) = for allt> iffx= ;
(iii) Nαx(t) =Nx(|αt|)for allx∈Xandα∈R\{};
(iv) Nx+y(s+t)≥T(Nx(s),Ny(t))for allx,y∈Xands,t≥.
In this case, the triplet (X,N,T) is said to be a probabilistic normed space.
In the above definition, forx∈X, the distribution functionN(x) is denoted byNxand
Nx(t) is the valueNxatt∈R.
Example . Let (X, · ) be a normed linear space. For allx∈X, define the mapping
Nx(t) =
ift≤,
t
t+x ift> .
Then (X,N,Tp) and (X,N,Tm) are probabilistic normed spaces.
Example . Let (X, · ) be a normed linear space. For allx∈X, define the mapping
Nx(t) =
ift≤,
e–tx ift> .
Then (X,N,Tp) is a probabilistic normed space.
Now, let us recall some topological properties of probabilistic normed spaces.
Definition . Let (X,N,T) be a probabilistic normed space. A sequence{xn}inXis said to be convergent to a pointx∈Xif for anyε> andλ> , there exists a positive integer
Nsuch that
Nxn–x(ε) > –λ
for everyn≥N.
Definition . Let (X,N,T) be a probabilistic normed space. A sequence{xn}inXis said to be Cauchy if for anyε> andλ> , there exists a positive integerNsuch that
Nxn–xm(ε) > –λ
Definition . Let (X,N,T) be a probabilistic normed space. It is said to be a Banach probabilistic normed space (or complete) if every Cauchy sequence inXis convergent to a point inX.
Definition . Let (X,N,T) be a probabilistic normed space. A subsetAofXis said to be closed if every convergent sequence inAconverges to an element ofA.
Definition . Let (X,N,T) be a probabilistic Banach space. A nonempty subsetP⊆X
is a cone if it satisfies the following conditions: (i) Pis closed and convex;
(ii) ifp∈P,tp∈Pfor everyt≥; (iii) if bothpand–pare inP, thenp= .
Letbe the partial order onXinduced by the conePinX. That is,
p,q∈X, pq ⇐⇒ p–q∈P.
ThusXbecomes a partially ordered probabilistic Banach space. Ifx,y∈X, the notation
x≺ymeans thatxyandx=y.
Definition . Let (X,N,T) be a probabilistic Banach space. A conePinXis said to be normal if there is some constantK> (normal constant) such that
x,y∈X, xy ⇒ Nx(t)≥Ny
t K
, t∈R.
Definition . Let (X,N,T) be a probabilistic Banach space andP⊆Xbe a cone inX. LetCbe a convex subset inX. An operatorA:C→Xis called a convex operator if
Atx+ ( –t)ytAx+ ( –t)Ay
for allx,y∈C,xyandt∈[, ].
Definition . Let (X,N,T) be a probabilistic Banach space andP⊆Xbe a cone inX. An operatorA:C→Xis said to be a decreasing operator if
x,y∈C, xy ⇒ AxAy.
2 Main result and proof
Before stating our main result, we need some lemmas.
Lemma . Let(X,N,T)be a probabilistic Banach space and{xn}be a sequence in X that converges to some x∈X.Then any subsequence of{xn}converges to x.
Proof Let{xϕ(n)}be a subsequence of{xn}, whereϕ:N→Nis a mapping satisfying
ϕ(n+ ) >ϕ(n)
for everyn∈N. Letε> andλ> . Since{xn}converges tox∈X, there is some positive integerNsuch that
Nxn–x(ε) > –λ
for everyn≥N. On the other hand,
n≥N ⇒ ϕ(n)≥ϕ(N)≥N. Then
Nxϕ(n)–x(ε) > –λ
for everyn≥N. This proves that{xϕ(n)}converges tox. Lemma .(see []) Let(X,N,T)be a probabilistic Banach space.Let{xn}and{yn}be two sequences in X and{αn}be a real sequence.The following properties hold:
(i) if{xn}converges tox∈Xand{yn}converges toy∈X,then{xn+yn}converges tox+y; (ii) if{αn}converges to someα∈Rand{xn}converges to somex∈X,then{αnxn}
converges toαx.
Lemma . Let(X,N,T)be a probabilistic Banach space and P⊆X be a normal cone in X with normal constant K> .Let{un}be a sequence in X such that
um–unξna
for every m,n≥N,where a∈X and{ξn}is a real sequence such thatξn→as n→ ∞. Then{un}is a Cauchy sequence in the probabilistic Banach space(X,N,T).
Proof Letε> andλ> . Without restriction of the generality, we may assume thatξn=
for everyn≥N. SincePis a normal cone, we have
Num–un(ε)≥Nξna
ε
K
=Na
ε
K|ξn|
(.)
for everym,n≥N. On the other hand, sinceNa∈D, we have
sup
Then there is somet∗∈Rsuch that
Nat∗> –λ.
Sinceξn→ asn→ ∞, there is some positive integerNsuch that
ε
K|ξn|
>t∗
for everyn≥N. SinceNais non-decreasing, we get
Na
ε
K|ξn|
≥Na
t∗> –λ (.)
for everyn≥N. Finally, using (.) and (.), we obtain
Num–un(ε) > –λ
for everyn,m≥max{N,N}. This proves that{un}is a Cauchy sequence in the
probabilis-tic Banach space (X,N,T).
Lemma . Let(X,N,T)be a probabilistic Banach space and P⊆X be a normal cone in X with normal constant K> .Let us consider two sequences{un}and{vn}in X such that
unvn for every n≥N.Then
{vn}converges to ⇒ {un}converges to.
Proof Letε> andλ> . Since{vn}converges to , there exists someN∈Nsuch that
Nvn
ε
K
> –λ
for anyn≥N. On the other hand, sincePis normal, we have
Nun(ε)≥Nvn
ε
K
for anyn≥N. Thus we proved that
Nun(ε) > –λ
for anyn≥max{N,N}, which implies that{un}converges to . The following result is an immediate consequence of Lemma . and Lemma ..
such that
unvnwn
for every n≥N.Then
{un},{wn}converge to∈X ⇒ {vn}converges to.
Now, we are ready to state and prove our main result.
Let (X,N,T) be a probabilistic Banach space andP⊆Xbe a normal cone inX with normal constant K> . We denote byAthe set of operatorsA:P→P satisfying the following conditions:
(A) ≺A;
(A) Ais a convex and decreasing operator;
(A) there existγ∈(, )andm,n∈Nwithn>msuch that
Am+ –AmγAm+ –Am (.) and
An
Am+ +Am. (.)
Theorem . Let A∈A.Then
(i) Ahas a unique fixed pointx∗∈P;
(ii) for any initial valuex∈P,the Picard sequence{xn}inXdefined by
xn=Axn–, n≥
converges tox∗; (iii) we have the estimates
Nx(m+n)–x∗(t)
≥T
NA
(n– –n+m)t
K
,NA
(n– –n+m)t
K
(.)
for everyn>n+ –m,t∈R,and
Nx(m+n)+–x∗(t)
≥T
NA
(n– –n+m)t
K
,NA
(n– –n+m)t
K
(.)
for everyn>n+ –m,t∈R.
Proof Let us consider the sequence{un}inPdefined by
By (.) and (.), we have
u(m+)γum++ ( –γ)um (.)
and
un–umum+. (.)
SinceAis a decreasing operator, we have
=uu · · · um · · · u(m+n) · · ·
u(m+n)+ · · · um+ · · · uu=A. (.)
Using (.) and (.), we obtain
γ(u(m+n)+–um) · · · γ(um+–um)
u(m+)–um · · · u(m+n)–um.
For everyn∈N, we define the set Sn=
s∈(, ] :u(m+n)su(m+n)++ ( –s)um
.
ClearlySn=∅sinceγ ∈Snfor everyn∈N. For everyn∈N, let sn=sup
R Sn and en= –sn.
Letn∈Nbe fixed. By the definition of sn, there exists a sequence {ap} ⊂Snsuch that ap→snasp→ ∞. Thus we have
u(m+n)apu(m+n)++ ( –ap)um for everyp∈N. This means that
u(m+n)–apu(m+n)+– ( –ap)um∈P
for everyp∈N. SincePis closed, lettingp→ ∞, using Lemma . and (.), we obtain
u(m+n)snu(m+n)++enum,
<γ ≤s≤s≤ · · · ≤sn≤ · · · ≤, ≤en≤ –γ. (.)
Now, we shall prove thaten→ asn→ ∞. Using inequalities (.), (.), (.) and the fact thatAis a decreasing convex operator, for everyn≥n–m– , we have
u(m+n)+=Au(m+n)A(snu(m+n)++enum)
snu(m+n+)+en(un–um)
( +en)u(m+n+)–enum.
Thus we proved that
u(m+n+)
+enu(m+n)++ en
+enum
for everyn≥n–m– . Using thatAis convex and decreasing and inequality (.), we obtain
u(m+n+)+=Au(m+n+) +en
u(m+n+)+
en
+en um+
+enu(m+n+)+ en
+en(un–um)
+ en
+en u(m+n+)– en
+enum,
which implies that
u(m+n+) +en
+ enu(m+n+)++ en
+ enum
for everyn≥n–m– . This implies that +en
+ en∈Sn+
for everyn≥n–m– . So, for everyn≥n–m– , we have
–en+=sn+≥ +en
+ en
,
that is,
en+≤
en
+ en.
Now, let us consider the functionf : [, ]→[, ] defined by
f(r) = r
+ r, r∈[, ].
Since
en+≤f(en)
for everyn≥n–m– andf is a non-decreasing function, we obtain ≤en+≤fn–n+m+(en–m–)
≤ en–m–
Thus we have
≤en+≤
(n–n+m)
(.)
for everyn≥n–m+ . Lettingn→ ∞in (.), we get
lim
n→∞en= ,
that is,
lim
n→∞sn= .
Using (.), we deduce that for alln,p∈N, we have u(m+n+p)–u(m+n)u(m+n)+–u(m+n)
en(u(m+n)+–um)en(u–um)enu. (.)
Since P is a normal cone and en→ as n→ ∞, using Lemma ., we obtain that
{u(m+n)}is a Cauchy sequence. SincePis closed and (X,N,T) is a complete probabilistic normed space, there isx∗∈Psuch that{un}converges tox∗. Using (.), Lemma . and
Lemma ., we deduce that
{un}and{un+}converge tox∗. (.)
On the other hand, using (.), we have
u(m+n)u(m+n+p)
for everyn,p∈N. This implies that
u(m+n+p)–u(m+n)∈P
for everyn,p∈N. Fixn∈Nand lettingp→ ∞, from (.) and sincePis closed, we get
x∗–u(m+n)∈P, that is,
u(m+n)x∗.
Similarly, we can observe that
x∗u(m+n)+.
Thus we proved that
for alln∈N. Using the fact thatAis a decreasing operator, (.) yields
u(m+n+)Ax∗u(m+n)+
for alln∈N. Lettingn→ ∞in the above inequality and using (.), we obtain
x∗Ax∗x∗,
that is,x∗=Ax∗. Thus we proved thatx∗∈Pis a fixed point of the operatorA.
Now, let x∈Pbe an arbitrary point. Let us consider the Picard sequence {xn} inP defined by
xn=Anx for everyn∈N. (.)
SinceAis a decreasing operator, using thatuxandux, by induction we obtain
u(m+n)x(m+n)u(m+n)–, (.)
u(m+n)x(m+n)+u(m+n)+ (.)
for everyn∈N. Lettingn→ ∞in (.) and (.), using (.) and Lemma ., we obtain
{xn}and{xn+}converge tox∗,
which implies that{xn}converges tox∗.
Let us prove thatx∗is the unique fixed point of the operatorA. Suppose thaty∗∈Pis another fixed point ofA. Letx=y∗ and consider the Picard sequence{xn}inPdefined by (.). Then we have
xn=Anx=y∗ for everyn∈N.
Lettingn→ ∞in (.) and (.), using (.), we obtainx∗=y∗. Thus we proved thatx∗
is the unique fixed point ofA.
Let us prove estimate (.). Lett∈R, we have
Nx(m+n)–x∗(t) =N(x(m+n)–u(m+n))+(u(m+n)–x∗)(t)
≥T
Nx(m+n)–u(m+n)
t
,Nu(m+n)–x∗
t
. (.)
By (.), we have
x(m+n)–u(m+n)u(m+n)––u(m+n).
SincePis a normal cone, we get
Nx(m+n)–u(m+n)
t
≥Nu(m+n)––u(m+n)
t
K
Using (.) and (.), we have
x∗–u(m+n)u(m+n)––u(m+n).
SincePis a normal cone, we get
Nx∗–u
(m+n)
t
≥Nu(m
+n)––u(m+n)
t
K
. (.)
It follows from (.), (.) and (.) that
Nx(m+n)–x∗(t)≥T
Nu(m+n)––u(m+n)
t
K
,Nu(m+n)––u(m+n)
t
K
. (.)
On the other hand, by (.) and (.), we have
u(m+n)u(m+n–)sn–u(m+n)–+en–um. Using (.), (.) and the above inequality, we obtain
u(m+n)––u(m+n)en–A.
Using (.) and the above inequality, we obtain
u(m+n)––u(m+n)
(n– –n+m)
A
for everyn>n+ –m. SincePis a normal cone, we have
Nu(m+n)––u(m+n)
t
K
≥NA
(n– –n+m)t
K
(.)
for everyn>n+ –m. Now, (.) follows immediately from (.) and (.). The proof of estimate (.) follows using similar arguments as above. This ends the proof.
3 Positive solutions for the nonlinear functional equation:x=x0+Bx
In this section, from our main theorem (Theorem .), we deduce an existence and uniqueness result for the nonlinear operator equation on ordered probabilistic Banach spaces
x=x+Bx, (.)
wherex∈PandB:P→Pis a given operator satisfying certain conditions. There have appeared a series of research results concerning this kind of nonlinear operator Eq. (.) because of the crucial role played by nonlinear equations in applied science as well as in mathematics (see [–]).
(B) B = ;
(B) Bis a convex and decreasing operator.
We have the following result.
Theorem . Let B∈Band x∈P such that≺x.Then the operator Eq. (.)has a
unique solution x∗∈P.
Proof Let us define the operatorA:P→Pby
Ax=x+Bx, x∈P.
Obviously,x∈Pis a solution to Eq. (.) if and only ifxis a fixed point ofA. We have just to prove that the operatorAsatisfies the required conditions of Theorem ., that is,A
belongs to the class of operatorsA.
•Condition (A). Sincex, using (B), we have
A =x+B =x. Thus condition (A) is satisfied.
• Condition (A). Let x,y∈P such thatxy. Using the fact thatBis a decreasing operator, we obtain
BxBy, which implies that
x+Bxx+By, that is,
AxAy.
ThusAis a decreasing operator.
Now, lett∈[, ],x,y∈Psuch thatxy. SinceBis a convex operator, we have
Btx+ ( –t)ytBx+ ( –t)By, which implies that
x+B
tx+ ( –t)yx+tBx+ ( –t)By=t(x+Bx) + ( –t)(x+By). Thus we have
Atx+ ( –t)ytAx+ ( –t)Ay.
•Condition (A). Letn= andm= . SinceB = , we have
An =A =A(x
+B) =Ax=x+Bx.
On the other hand,
Am+ +Am= A =
x.
Clearly, we have
An
Am+ +Am.
Now, we have
Am+ –Am =A =x +Bx.
Moreover, for anyγ ∈(, ), we have
γAm+ –Am=γA =γx
+γB(x+Bx).
Thus we have
Am+ –Am–γAm+ –Am= ( –γ)x
+Bx–γB(x+Bx).
Sinceγ∈(, ) andBis a decreasing operator, we have ( –γ)x
and
Bx–γB(x+Bx)Bx–B(x+Bx). Then we get
Am+ –AmγAm+ –Am for everyγ ∈(, ).
HenceA∈Aand the result follows from Theorem ..
4 The case of Banach spaces
Let (X, · ) be a Banach space andP⊂Xbe a cone inX. Let us suppose thatPis a normal cone with normal constantK> , that is,
x,y∈X, xy ⇒ x ≤Ky.
(i) ≺A;
(ii) Ais a convex and decreasing operator;
(iii) there existγ∈(, )andm,n∈Nwithn>msuch that
Am+ –AmγAm+ –Am
and
An
Am+ +Am.
Let us consider the probabilistic Banach space (X,N,Tm), where
Nx(t) =
ift≤,
t
t+x ift>
for everyx∈X.
Lemma . P is a normal cone in the probabilistic Banach space(X,N,Tm)with normal
constant K.
Proof At first, let us prove thatPis also a cone in the probabilistic Banach space (X,N,Tm). Indeed, we have just to prove thatPis also closed in (X,N,Tm). Let{xn}be a sequence in Psuch that{xn}converges to somex∈Xin (X,N,Tm). Letε> , by the definition of the convergence in a probabilistic normed space, there exists someN∈Nsuch that
Nxn–x(ε) >
for everyn≥N. Then we have
ε ε+xn–x>
for everyn≥N, which is equivalent to
xn–x<ε
for everyn≥N. Thus {xn}converges toxwith respect to · . SincePis closed with respect to the topology of the norm · , we havex∈P. Then we have proved thatPis also closed in the probabilistic Banach space (X,N,Tm).
Now, let us prove thatPis normal in (X,N,Tm). Letx,y∈Xsuch that
xy.
SincePis a normal cone in the Banach space (X, · ) with normal constantK, we have
Then, for everyt> , we have
Nx(t) = t t+x≥
t t+Ky=
t K t K +y
=Ny
t K
.
Ift≤, obviously we have
=Nx(t) =Ny
t K
.
Thus, for everyt∈R, we have
Nx(t)≥Ny
t K
.
This proves thatPis a normal cone in (X,N,Tm) with normal constantK.
Now, using Theorem . and Lemma ., we obtain the following fixed point result in Banach spaces ([], Theorem .).
Corollary . Suppose that conditions(i)-(iii)are satisfied.Then
(I) Ahas a unique fixed pointx∗∈P;
(II) for any initial valuex∈P,the Picard sequence{xn}inXdefined by
xn=Axn–, n≥
converges tox∗; (III) we have the estimates
x(m+n)–x∗ ≤
KA
n– –n+m
for everyn>n+ –m,and
x(m+n)+–x∗ ≤
KA
n– –n+m
for everyn>n+ –m.
5 An application to a two-point boundary value problem
In this section, we present an application of Theorem . to a two-point boundary value problem.
Leta: [, ]→Rbe a given function satisfying the following conditions: (a) ais a continuous function;
(a) a(x( –x)) =a(x)for everyx∈[, ]; (a) <m≤a(x)≤Mfor everyx∈[, ].
Example . Leta: [, ]→Rbe a positive constant function. Thenasatisfies conditions (a)-(a).
Example . Leta: [, ]→Rbe the function defined by
a(x) =α+βx( –x), x∈[, ],
whereα> andβ≥ are constants. Thenasatisfies conditions (a)-(a) with
m=α and M=α+β
.
Letf : [,∞)→Rbe a function satisfying the following conditions: (f) f is a continuous function,f ≥;
(f) f is a decreasing and convex function; (f) f() = , <f(M) < ;
(f) f(γx)≥ for everyx∈[,M], where
γ = – f
M
–f
m
.
The set of functionsf: [,∞)→Rsatisfying the above conditions is not empty. Example . Letm=M= , that is,
a(x) = , x∈[, ].
Letf : [,∞)→Rbe the function defined by
f(x) =
+ x, x≥.
Thenf satisfies conditions (f)-(f) withγ =,,.
Now, let us consider the following two-point boundary value problem:
–u(x) =f(a(x)u(x)) if <x< ,
u() =u() = . (.)
Let (X,N,Tm) be the probabilistic Banach space, where X=C([, ]) is the set of real
continuous functions in [, ] andN:X→Dis given by
Nu(t) =
ift≤,
t
t+max≤x≤|u(x)| ift> ,
u∈C[, ].
Let
ThenPis a normal cone in the probabilistic Banach space (X,N,Tm). The partial order induced by the conePin the setXis defined by
u,v∈C[, ], uv ⇐⇒ u(x)≤v(x) for allx∈[, ]. We have the following result.
Theorem . The boundary value problem(.)has a unique positive solution u∗∈P.
Proof The Green function associated to (.) is given by
G(x,y) =
y( –x) if ≤y≤x≤,
x( –y) if ≤x≤y≤.
Then problem (.) is equivalent to the integral equation
u(x) =
G(x,y)fa(y)u(y)dy, x∈[, ].
Let us consider the nonlinear operatorA:P→Pdefined by
(Au)(x) =
G(x,y)fa(y)u(y)dy, x∈[, ].
We have to prove thatAhas a unique fixed point inP. Theorem . will be used for the proof.
Clearly, the operatorAis convex and decreasing with respect to the partial order. For everyx∈[, ], we have
(A)(x) =
G(x,y)f()dy=
G(x,y)dy=x( –x) ≥. Then we have
≺A. Moreover, we have
≤a(x)(A)(x)≤M
, x∈[, ], which implies that
fa(x)(A)(x)≥f
M
, x∈[, ].
The above inequality yields
A(x) =
G(x,y)fa(y)(A)(y)dy≥f
M
On the other hand, for everyx∈[, ], we have
A(x) =
G(x,y)fa(y)(A)(y)dy
= ( –x) x
yf
a(y)y( –y)
dy+x
x
( –y)f
a(y)y( –y)
dy.
Using the fact thatf is a decreasing and convex function, we obtain
( –x) x
yf
a(y)y( –y)
dy
≤( –x) x
y( –y)f
a(y)y
dy+ ( –x) x
ydy
≤( –x) x
y( –y)f
a(y)
dy+ ( –x) x
y( –y)dy+ ( –x) x
ydy
≤( –x)f
m
x
y( –y)dy+ ( –x) x
y( –y)dy+ ( –x) x
ydy
= f m
x– x+ x– x+ x
(A)(x).
Similarly, using condition (a), we have
x
x
( –y)f
a(y)y( –y) dy ≤ f m
+x– x+ x+ – x+ x– x
(A)(x).
Thus we have
A(x)≤
+f
m
(A)(x) (.)
for everyx∈[, ]. Moreover, using (.), we get
fa(x)A(x)≤f
a(x)f
M
(A)(x)
(.)
for everyx∈[, ]. Now, (.) and (.) yield
A(x) =
G(x,y)fa(y)A(y)dy
≤
G(x,y)f
a(y)f
M
(A)(y) dy ≤f M
G(x,y)fa(y)(A)(y)dy+
–f
M
(A)(x)
=f
M
A(x) +
–f
M
(A)(x)
for everyx∈[, ]. Thus we have
fa(x)A(x)≥fγa(x)(A)(x), x∈[, ].
Using condition (f), we obtain
A(x) =
G(x,y)fa(y)A(y)dy
≥
G(x,y)fγa(y)(A)(y)dy
≥
(A)(x)
for everyx∈[, ]. Finally, using (.), for allx∈[, ], we have
A(x)≥(A)(x)≥A(x), where=f(M)∈(, ).
Now, the desired result follows from Theorem . withm= andn= .
Competing interests
The authors declare that they have no competing interests. Authors’ contributions
The authors contributed equally and significantly in writing this paper. Both authors read and approved the final manuscript.
Acknowledgements
This project was funded by the National Plan for Science, Technology and Innovation (MAARIFAH), King Abdulaziz City for Science and Technology, Kingdom of Saudi Arabia, Award Number (12-MAT 2913-02).
Received: 28 February 2015 Accepted: 22 May 2015 References
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