# Fixed Point Theorems in Ordered Banach Spaces and Applications to Nonlinear Integral Equations

Authors: Ravi P. Agarwal; Nawab Hussain; Mohamed-Aziz Taoudi
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We present some new common fixed point theorems for a pair of nonlinear mappings defined on an ordered Banach space. Our results extend several earlier works. An application is given to show the usefulness and the applicability of the obtained results.


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## Body

## 1. Introduction

The problem of existence of common fixed points to a pair of nonlinear mappings is now a classical theme. The applications to differential and integral equations made it more interesting. A considerable importance has been attached to common fixed point theorems in ordered sets [1, 2]. In a recent paper, Dhage [3] proved some common fixed point theorems for pairs of condensing mappings in an ordered Banach space. More recently, Hussain et al. [4] extended the results of Dhage to 1-set contractive mappings. In the present paper we pursue the investigations started in the aforementioned papers and prove some new common fixed point theorems under weaker assumptions. Also we present some common fixed point results using the weak topology of a Banach space. The use of the concepts of ws-compactness and ww-compactness increases the usefulness of our results in many practical situations especially when we work in nonreflexive Banach spaces. We will illustrate this fact by proving the existence of nonnegative integrable solutions for an implicit integral equation. For the remainder of this section we gather some notations and preliminary facts. Let X be a real Banach space, ordered by a cone K. A cone K is a closed convex subset of X with λK⊆K(λ≥0), and K∩(-K)={0}. As usual x≤y⇔y-x∈K.Definition 1.1.
LetX be an ordered Banach space with order ≤. A mapping T:D(T)⊆X→X is said to be isotone increasing if for all x,y∈D(T), we have that x≤y implies Tx≤Ty.Definition 1.2.
LetM be a nonempty subset of an ordered Banach space X with order ≤. Two mappings S,T:M→M are said to be weakly isotone increasing if Sx≤TSx and Tx≤STx hold for all x∈M. Similarly, we say that S and T are weakly isotone decreasing if Tx≥STx and Sx≥TSx hold for all x∈M. The mappings S and T are said to be weakly isotone if they are either weakly isotone increasing or weakly isotone decreasing.In our considerations the following definition will play an important role. Letℬ(X) denote the collection of all nonempty bounded subsets of X and 𝒲(X) the subset of ℬ(X) consisting of all weakly compact subsets of X. Also, let Br denote the closed ball centered at 0 with radius r.Definition 1.3 (see  [5]).
A functionψ:ℬ(X)→ℝ+ is said to be a measure of weak noncompactness if it satisfies the following conditions.(1)
The familyker(ψ)={M∈ℬ(X):ψ(M)=0} is nonempty and ker(ψ) is contained in the set of relatively weakly compact sets of X.(2)
M1⊆M2⇒ψ(M1)≤ψ(M2).(3)
ψ(co¯(M))=ψ(M), where co¯(M) is the closed convex hull of M.(4)
ψ(λM1+(1-λ)M2)≤λψ(M1)+(1-λ)ψ(M2) for λ∈[0,1].(5)
If(Mn)n≥1 is a sequence of nonempty weakly closed subsets of X with M1bounded and M1⊇M2⊇⋯⊇Mn⊇⋯ such that limn→∞ψ(Mn)=0,then M∞:=⋂n=1∞‍Mn is nonempty.The familykerψ described in (1) is said to be the kernel of the measure of weak noncompactness ψ. Note that the intersection set M∞ from (5) belongs to kerψ since ψ(M∞)≤ψ(Mn) for every n and limn→∞ψ(Mn)=0. Also, it can be easily verified that the measure ψ satisfies
(1.1)ψ(Mw¯)=ψ(M),
where Mw¯ is the weak closure of M.A measure of weak noncompactnessψ is said to be regular if
(1.2)ψ(M)=0iffMisrelativelyweaklycompact,subadditive if
(1.3)ψ(M1+M2)≤ψ(M1)+ψ(M2),homogeneous if
(1.4)ψ(λM)=|λ|ψ(M),λ∈R,
and set additive (or have the maximum property) if
(1.5)ψ(M1∪M2)=max⁡⁡(ψ(M1),ψ(M2)).The first important example of a measure of weak noncompactness has been defined by De Blasi [6] as follows:
(1.6)w(M)=inf⁡⁡{r>0:thereexistsW∈W(X)withM⊆W+Br},
for each M∈ℬ(X).Notice thatw(·) is regular, homogeneous, subadditive, and set additive (see [6]).By a measure of noncompactness on a Banach spaceX we mean a map ψ:ℬ(X)→ℝ+ which satisfies conditions (1)–(5) in Definition 1.3 relative to the strong topology instead of the weak topology. The concept of a measure of noncompactness was initiated by the fundamental papers of Kuratowski [7] and Darbo [8]. Measures of noncompactness are very useful tools in nonlinear analysis [9] especially the so-called Kuratowski measure of noncompactness [7] and Hausdorff (or ball) measure of noncompactness [10].Definition 1.4.
LetX be a Banach space and ψ a measure of (weak) noncompactness on X. Let A:D(A)⊆X→X be a mapping. If A(D(A)) is bounded and for every nonempty bounded subset M of D(A) with ψ(M)>0, we have, ψ(A(M)))<ψ(M), then A is called ψ-condensing. If there exists k, 0≤k≤1, such that A(D(A)) is bounded and for each nonempty bounded subset M of D(A), we have ψ(A(M))≤kψ(M), then A is called k-ψ-contractive.Remark 1.5.
Clearly, everyk-ψ-contractive map with k<1 is ψ-condensing and every ψ-condensing map is 1-ψ-contractive.Definition 1.6 (see [11]).
A mapA:D(A)→X is said to be ws-compact if it is continuous and for any weakly convergent sequence (xn)n∈ℕ in D(A) the sequence (Axn)n∈ℕ has a strongly convergent subsequence in X.Remark 1.7.
The concept of ws-compact mappings arises naturally in the study of both integral and partial differential equations (see [11–19]).Definition 1.8.
A mapA:D(A)→X is said to be ww-compact if it is continuous and for any weakly convergent sequence (xn)n∈ℕ in D(A) the sequence (Axn)n∈ℕ has a weakly convergent subsequence in X.Definition 1.9.
LetX be a Banach space. A mapping T:D(T)⊆X→X is called a nonlinear contraction if there exists a continuous and nondecreasing function φ:ℝ+→ℝ+ such that
(1.7)‖Tx-Ty‖≤φ(‖x-y‖)
for all x,y∈D(T), where φ(r)<r for r>0.Remark 1.10.
It is easy to prove that a nonlinear contraction mapping is a nonlinear set-contraction with respect to the Kuratowskii measure of noncompactness. We will prove that the same property holds for the De Blasi measure of weak noncompactness provided that the nonlinear contraction mapping is ww-compact.Lemma 1.11.
LetT be a ww-compact nonlinear contraction mapping on a Banach X. Then for each bounded subset M of X one has
(1.8)w(TM)≤ϕ(w(M)).
Here, w is the De Blasi measure of weak noncompactness.Proof.
LetM be a bounded subset of X and r>w(M). There exist 0≤r0<r and a weakly compact subset W of X such that M⊆W+Br0. Since T is a ϕ-nonlinear contraction, then
(1.9)TM⊆TW+Bϕ(r0)⊆TW¯w+Bϕ(r0).
Moreover, since T is ww-compact, then TW¯w is weakly compact. Accordingly,
(1.10)w(TM)≤ϕ(r0)≤ϕ(r).
Letting r→w(M) and using the continuity of ϕ we deduce that
(1.11)w(TM)≤ϕ(w(M)).The following theorem is a sharpening of [3, Theorem 2.1] and [4, Theorem 3.1].Theorem 1.12.
LetX be an ordered Banach space and ψ a set additive measure of noncompactness on X. Let M be a nonempty closed convex subset of X and S,T:M→M be two continuous mappings satisfying the following: (i)
T is 1-ψ-contractive,(ii)
S is ψ-condensing,(iii)
S and T are weakly isotone.Then T and S have a common fixed point.Proof.
Letx∈M be fixed. Consider the sequence {xn} defined by
(1.12)x0=x,x2n+1=Sx2n,x2n+2=Tx2n,n=0,1,2,….
Suppose first that S and T are weakly isotone increasing on M. Then from (1.12) it follows that
(1.13)x1≤x2≤⋯≤xn≤⋯.
Set A={x1,x2,…,xn,…},A1={x2,x4,…,x2n,…}, and A2={x1,x3,…,x2n+1,…}. Clearly
(1.14)A=A1∪A2,A2=S(A1)∪{x1},A1=T(A2).
We show that A1 is relatively compact. Suppose the contrary, then ψ(A1)>0. Since T is 1-ψ-contractive and S is ψ-condensing, it follows from (1.14) that
(1.15)ψ(A2)=ψ(S(A1)∪{x1})=ψ(S(A1))<ψ(A1),ψ(A1)=ψ(T(A2))≤ψ(A2).
This is impossible. Thus A1 is relatively compact. The continuity of S implies that A2=S(A1)∪{x1} is relatively compact. Consequently A is relatively compact. Since {xn} is monotone increasing in A, then it is convergent. Let x* be its limit. Using the continuity of S and T we get
(1.16)x*=lim⁡n→∞x2n+1=lim⁡n→∞S(x2n)=S(lim⁡n→∞x2n)=S(x*),x*=lim⁡n→∞x2n+2=lim⁡n→∞T(x2n+1)=T(lim⁡n→∞x2n+1)=T(x*).
To complete the proof we consider the case where S and T are weakly isotone decreasing on M. In this case, the sequence {xn} is monotone decreasing and then converges to a common fixed point of S and T.Remark 1.13.
In [3, Theorem 2.1] T is assumed to be set condensing, while S is assumed to be affine, X is assumed to be reflexive and I-T is demiclosed in [4, Theorem 3.1].Remark 1.14.
As an application of Theorem1.12, the modified versions of Theorems 3.6–3.12 in [4] can be proved similarly.As easy consequences of Theorem1.12 we obtain the following results.Corollary 1.15.
LetM be a nonempty closed convex subset of a Banach space X and let S,T:M→M be two completely continuous mappings. Also assume T and S are weakly isotone mappings. Then T and S have a common fixed point.Corollary 1.16.
LetM be a nonempty closed convex subset of a Banach space X and S,T:X→X be two continuous mappings. Assume T is nonexpansive, that is, ∥Tx-Ty∥≤∥x-y∥ for all x,y∈X, and S is a nonlinear contraction. Also assume that S and T are weakly isotone mappings, T(M)⊆M and S(M)⊆M. Then T and S have a common fixed point in M.Corollary 1.17.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of noncompactness on X. Let S,T:X→X be two continuous mappings satisfying the following:  (i)
T(M)⊆M and S(M)⊆M,(ii)
T is a nonexpansive mapping,(iii)
S is ψ-condensing,(iv)
T and S are weakly isotone.Then S and T have a common fixed point in M.Note that in Corollary1.17 we do not have uniqueness. However, if we assume T is shrinking, we obtain uniqueness. Recall that T:D(T)⊆X→X is shrinking if
(1.17)‖Tx-Ty‖<‖x-y‖wheneverx,y∈D(T)withx≠y.Corollary 1.18.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of noncompactness on X. Let S,T:X→X be two continuous mappings satisfying the following:  (i)
T(M)⊆M and S(M)⊆M,(ii)
T is a shrinking mapping,(iii)
S is ψ-condensing,(iv)
T and S are weakly isotone.Then S and T have a unique common fixed point in M.Proof.
From Corollary1.17 it follows that there exists x∈M such that Sx=Tx=x. Now from (1.17) we infer that T cannot have two different fixed points. This implies that S and T have a unique common fixed point.Corollary 1.19.
LetM be a nonempty closed convex subset of a Banach space X and S,T:X→X be two mappings satisfying the following:  (i)
T(M)⊆M and S(M)⊆M,(ii)
T is a shrinking mapping,(iii)
S is a nonlinear contraction,(iv)
T and S are weakly isotone.Then S and T have a unique common fixed point in M which is the unique fixed point of S.Theorem 1.20.
LetX be an ordered Banach space and ψ a set additive measure of weak noncompactness on X. Let M be a nonempty closed convex subset of X and S,T:M→M be two sequentially weakly continuous mappings satisfying the following:  (i)
T is 1-ψ-contractive,(ii)
S is ψ-condensing,(iii)
S and T are weakly isotone.Then T and S have a common fixed point in M.Proof.
Letx∈M be fixed. Consider the sequence {xn} defined by
(1.18)x0=x,x2n+1=Sx2n,x2n+2=Tx2n,n=0,1,2,….
Suppose first that S and T are weakly isotone increasing on M. Then from (1.18) it follows that
(1.19)x1≤x2≤⋯≤xn≤⋯.
As in the proof of Theorem 1.12 we set A={x1,x2,…,xn,…},  A1={x2,x4,…,x2n,…}, and A2={x1,x3,…,x2n+1,…}. Clearly
(1.20)A=A1∪A2,A2=S(A1)∪{x1},A1=T(A2).
We show that A1 is relatively weakly compact. Suppose the contrary, then ψ(A1)>0. Since T is 1-ψ-contractive and S is ψ-condensing, it follows from (1.20) that
(1.21)ψ(A2)=ψ(S(A1)∪{x1})=ψ(S(A1))<ψ(A1),ψ(A1)=ψ(T(A2))≤ψ(A2).
This is impossible. Thus A1 is relatively weakly compact. The weak sequential continuity of S implies that A2=S(A1)∪{x1} is relatively weakly compact. The same reasoning as in the proof of Theorem 1.12 gives the desired result.As easy consequences of Theorem1.20 we obtain the following results.Corollary 1.21.
LetM be a nonempty closed convex subset of a Banach space X and let S,T:M→M be two weakly completely continuous mappings. Also assume T and S are weakly isotone mappings. Then T and S have a common fixed point in M.Corollary 1.22.
LetM be a nonempty closed convex subset of a Banach space X and let S,T:X→X be two sequentially weakly continuous mappings. Assume T is nonexpansive and S is a nonlinear contraction. Also assume that S and T are weakly isotone mappings, T(M)⊆M and S(M)⊆M. Then T and S have a common fixed point in M.Corollary 1.23.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of weak noncompactness on X. Let S,T:X→X be two sequentially weakly continuous mappings satisfying the following:  (i)
T is a nonexpansive mapping,(ii)
S is ψ-condensing,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a common fixed point.Proof.
This follows from Theorem1.20 on the basis of Lemma 1.11.Corollary 1.24.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of weak noncompactness on X. Let S,T:X→X be two sequentially weakly continuous mappings satisfying the following:  (i)
T is a shrinking mapping,(ii)
S is ψ-condensing,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a unique common fixed point in M.Corollary 1.25.
LetM be a nonempty closed convex subset of a Banach space X and let S,T:X→X be two sequentially weakly continuous mappings satisfying the following:  (i)
T is a shrinking mapping,(ii)
S is a nonlinear contraction,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a unique common fixed point in M which is the unique fixed point of S.Remark 1.26.
It is worth noting that, in some applications, the weak sequential continuity is not easy to be verified. The ws-compactness seems to be a good alternative (see [16] and the references therein). In the following we provide a version of Theorem 1.20 where the weak sequential continuity is replaced with ws-compactness.Theorem 1.27.
LetX be an ordered Banach space and ψ a set additive measure of weak noncompactness on X. Let M be a nonempty closed convex subset of X and S,T:M→M two continuous mappings satisfying the following:  (i)
T is 1-ψ-contractive,(ii)
S is ws-compact and ψ-condensing,(iii)
S and T are weakly isotone.Then T and S have a common fixed point.Proof.
Letx∈M be fixed. Consider the sequence {xn} defined by
(1.22)x0=x,x2n+1=Sx2n,x2n+2=Tx2n,n=0,1,2,….
Suppose first that S and T are weakly isotone increasing on M. Then from (1.22) it follows that
(1.23)x1≤x2≤⋯≤xn≤⋯.
Set A={x1,x2,…,xn,…},A1={x2,x4,…,x2n,…}, and A2={x1,x3,…,x2n+1,…}. Clearly
(1.24)A=A1∪A2,A2=S(A1)∪{x1},A1=T(A2).
Similar reasoning as in Theorem 1.20 gives that A1 is relatively weakly compact. From the ws-compactness of S it follows that A2=S(A1)∪{x1} is relatively compact. Now the continuity of T yields that A1=T(A2) is relatively compact. Consequently, A=A1∪A2 is relatively compact. The rest of the proof runs as in the proof of Theorem 1.12.As easy consequences of Theorem1.27 we obtain the following results.Corollary 1.28.
LetS,T:X→X be two continuous mappings. Assume T is nonexpansive and ww-compact and S is a ws-compact nonlinear contraction. Also assume that S and T are weakly isotone mappings. Then T and S have a common fixed point.Proof.
This follows from Theorem1.27 on the basis of Lemma 1.11.Corollary 1.29.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of weak noncompactness on X. Let S,T:X→X be two continuous mappings satisfying the following the following:  (i)
T is a ww-compact nonexpansive mapping,(ii)
S is ws-compact and ψ-condensing,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a common fixed point in M.Proof.
This follows From Theorem1.27 on the basis of Lemma 1.11.Corollary 1.30.
LetM be a nonempty closed convex subset of a Banach space X and ψ a set additive measure of weak noncompactness on X. Let S,T:X→X be two continuous mappings satisfying the following:  (i)
T is a ww-compact shrinking mapping,(ii)
S is ws-compact and ψ-condensing,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a unique common fixed point in M.Corollary 1.31.
LetM be a nonempty closed convex subset of a Banach space X and let S,T:X→X two continuous mappings satisfying the following:  (i)
T is a ww-compact and shrinking mapping,(ii)
S is a ws-compact nonlinear contraction,(iii)
T and S are weakly isotone,(iv)
T(M)⊆M and S(M)⊆M.Then S and T have a unique common fixed point in M which is the unique fixed point of S.Note that ifX is a Banach Lattice, then the ws-compactness in Theorem 1.27 can be removed, as it is shown in the following result.Theorem 1.32.
LetX be a Banach lattice and ψ a set additive measure of weak noncompactness on X. Let M be a nonempty closed convex subset of X and S,T:M→M two continuous mappings satisfying the following:  (i)
T is 1-ψ-contractive,(ii)
S is ψ-condensing,(iii)
S and T are weakly isotone.Then T and S have a common fixed point in M.Proof.
Letx∈M be fixed. Consider the sequence {xn} defined by
(1.25)x0=x,x2n+1=Sx2n,x2n+2=Tx2n,n=0,1,2,….
Suppose first that S and T are weakly isotone increasing on M. Then from (1.25) it follows that
(1.26)x1≤x2≤⋯≤xn≤⋯.
Set A={x1,x2,…,xn,…},A1={x2,x4,…,x2n,…}, and A2={x1,x3,…,x2n+1,…}. Clearly
(1.27)A=A1∪A2,A2=S(A1)∪{x1},A1=T(A2).
Similar reasoning as in Theorem 1.20 gives that A1 is relatively weakly compact. Thus {xn} has a weakly convergent subsequence, say {xnk}. Since in a Banach lattice every weakly convergent increasing sequence is norm-convergent we infer that {xnk} is norm-convergent. Consequently A1 is relatively compact. Similar reasoning as in the proof of Theorem 1.12 gives the desired result.


## 2. Application to Implicit Integral Equations

The purpose of this section is to study the existence of integrable nonnegative solutions of the integral equation given by(2.1)p(t,x(t))=∫01‍ζ(t,s)f(s,x(s))ds,t∈[0,1].Integral equations like (2.1) were studied in [20] in L2[0,1] and in [4] in Lp[0,1] with 1<p<+∞. In this section, we look for a nonnegative solution to (2.1) in L1[0,1]. For the remainder we gather some definitions and results from the literature which will be used in the sequel. Recall that a function f:[0,1]×ℝ→ℝ is said to be a Carathéodory function if (i)
for any fixedu∈ℝ, the function t→f(t,u) is measurable from [0,1] to ℝ,(ii)
for almost anyx∈[0,1], the function f(t,·):ℝ→ℝ is continuous.Letm[0,1] be the set of all measurable functions x:[0,1]→ℝ. If f is a Carathéodory function, then f defines a mapping Nf:m[0,1]→m[0,1] by Nf(x)(t):=f(t,x(t)). This mapping is called the superposition (or Nemytskii) operator generated by f. The next two lemmas are of foremost importance for our subsequent analysis.Lemma 2.1 (see [21, 22]).
Letf:[0,1]×ℝ→ℝ be a Carathéodory function. Then the superposition operator Nf maps continuously L1[0,1] into itself if and only if there exist a constant b≥0 and a function a(·)∈L+1[0,1] such that
(2.2)|f(t,u)|≤a(t)+b|u|,
where L+1[0,1] denotes the positive cone of the space L1[0,1].Lemma 2.2 (see [14]).
Iff:[0,1]×ℝ→ℝ is a Carathéodory function and Nf maps continuously L1[0,1] into itself, then Nf is ww-compact.Remark 2.3.
Although the Nemytskii operatorNf is ww-compact, generally it is not weakly continuous. In fact, only linear functions generate weakly continuous Nemytskii operators in L1 spaces (see, for instance, [23, Theorem 2.6]).The problem of existence of nonnegative integrable solutions to (2.1) will be discussed under the following assumptions: (a)
ζ:[0,1]×[0,1]→ℝ is strongly measurable and ∫01‍ζ(·,s)y(s)ds∈L1[0,1] whenever y∈L1[0,1] and there exists a function θ:[0,1]→ℝ belonging to L∞[0,1] such that 0≤ζ(t,s)≤θ(t) for all (t,s)∈[0,1]×[0,1]. The function f:[0,1]×ℝ→ℝ is a Carathéodory function and there exist a constant b>0 and a function a(·)∈L+1[0,1] such that |f(t,u)|≤a(t)+b|u| for all t∈[0,1] and u∈ℝ. Moreover, f(t,x(t))≥0 whenever x∈L+1[0,1];(b)
the functionp:[0,1]×ℝ→ℝ is nonexpansive with respect to the second variable, that is, |p(t,u)-p(t,v)|≤|u-v| for all t∈[0,1] and u,v∈ℝ;(c)
for allr>0 and for all x∈Br+ we have
(2.3)0≤θ(t)‖a‖+bθ(t)∫01‍p(s,x(s))ds≤p(t,x(t))≤x(t),
where Br+={x∈L1[0,1]:0≤x(t)≤rforallt∈[0,1]},(d)
b∥θ∥∞<1.Remark 2.4.
Ifp(t,0)=0, then from assumption (b) it follows that p(t,x(t))≤x(t) whenever x∈L+1[0,1].Remark 2.5.
From assumption (b) it follows that(2.4)|p(t,x(t))|≤|p(t,0)|+|x(t)|.
Using Lemma 2.2 together with Lemma 2.1 we infer that 𝒩p is ww-compact.Theorem 2.6.
Assume that the conditions (a–d) are satisfied. Then the implicit integral equation (2.1) has at least one nonnegative solution in L1[0,1].Proof.
The problem (2.1) may be written in the shape:
(2.5)Tx=Sx,
where T and S are the nonlinear operators given by
(2.6)(Tx)(t)=p(t,x(t)),(Sx)(t)=∫01‍ζ(t,s)f(s,x(s)).
Note thatT and S may be written in the shape:
(2.7)T=Np,S=KNf.
Here𝒩p and 𝒩f are the superposition operators associated to the functions p and f, respectively, and K denotes the linear integral defined by
(2.8)K:L1[0,1]⟶L1[0,1]:u(t)⟼Ku(t):=μ∫01‍ζ(t,s)u(s)ds.
Letr0 be the real defined by
(2.9)r0=‖a‖‖θ‖1-b‖θ‖
and M be the closed convex subset of L1[0,1] defined by
(2.10)M=Br0+={x∈L1[0,1]:0≤x(t)≤r0∀t∈[0,1]}.
Note that, for anyx∈L1[0,1], the functions Tx and Sx belong to L1[0,1] which is a consequence of the assumptions (a, b). Moreover, from assumption (c) it follows that for each x∈M we have
(2.11)0≤p(t,x(t))≤x(t)≤r0.
This implies that T(M)⊆M. Also, in view of assumption (b) we have for all x∈M and for all s∈[0,1](2.12)0≤f(s,x(s))≤a(s)+bx(s)≤a(s)+br0.
Hence
(2.13)0≤ζ(t,s)f(s,x(s))≤θ(t)(a(s)+br0).
Thus
(2.14)0≤∫01‍ζ(t,s)f(s,x(s))ds≤θ(t)(‖a‖+br0)≤‖θ‖∞(‖a‖+br0)=r0.
Therefore, S(M)⊆M.Our strategy consists in applying Corollary1.29 to find a nonnegative common fixed point for S and T in M. For the sake of simplicity, the proof will be displayed into four steps.Step 1.
S and T are continuous. Indeed, the assumption (a) and Lemma 2.1 guarantee that Np and 𝒩f map continuously L1[0,1] into itself. To complete the proof it remains only to show that K is continuous. This follows immediately from the hypothesis (b).Step 2.
S is ws-compact. Indeed, let (ρn) be a weakly convergent sequence of L1[0,1]. Using Lemma 2.2, the sequence (𝒩f(ρn)) has a weakly convergent subsequence, say (𝒩f(ρnk)). Let ρ be the weak limit of (𝒩f(ρnk)). Accordingly, keeping in mind the boundedness of the mapping ζ(t,·) we get(2.15)
∫
0
1
‍
ζ
(
t
,
s
)
f
(
s
,
ρ
n
k
(
s
)
)
d
s
→
∫
0
1
‍
ζ
(
t
,
s
)
f
(
s
,
ρ
(
s
)
)
d
s
.The use of the dominated convergence theorem allows us to conclude that the sequence(Sρnk) converges in L1[0,1].Step 3.
S maps bounded sets of L1[0,1] into weakly compact sets. To see this, let 𝒪 be a bounded subset of L1[0,1] and let C>0 such that ∥x∥≤C for all x∈𝒪. For x∈𝒪 we have
(2.16)|(Sx)(t)|dt≤∫01‍|ζ(t,s)||f(s,x(s))|ds≤∫01‍θ(t)(a(s)+b|x(s)|)ds≤θ(t)(‖a‖+bC).
Consequently,(2.17)∫E‍|(Sx)(t)|dt≤(‖a‖+bC)∫E‍θ(t)dt
for all measurable subsets E of [0,1]. Taking into account the fact that any set consisting of one element is weakly compact and using Corollary 11 in [24, page 294] we get lim|E|→0∫E‍|θ(t)|dt=0, where |E| is the Lebesgue measure of E. Applying Corollary 11 in [24, page 294] once again we infer that the set S(𝒪) is weakly compact.Step 4.
S and T are weakly isotone decreasing. Indeed, using assumption (c) we get
(2.18)∫01‍ζ(t,s)f(s,x(s))ds≥p(t,∫01‍ζ(t,s)f(s,x(s))ds),
and from our assumptions we have
(2.19)∫01‍ζ(t,s)f(s,p(s,x(s)))ds≤∫01‍ζ(t,s)a(s)ds+b∫01‍ζ(t,s)p(s,x(s))ds≤θ(t)‖a‖+bθ(t)∫01‍p(s,x(s))ds≤p(t,x(t)).
Hence Sx≥TSx and Tx≥STx and therefore S and T are weakly isotone decreasing.Note that theSteps1–4 show that the hypotheses of Corollary 1.29 are satisfied. Thus S and T have a common fixed point and therefore the problem (2.1) has at least one nonnegative solution in M.



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*Source: 245872-2012-07-15.xml*