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

and Applied Analysis 3


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 wscompactness 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.

Abstract and Applied Analysis
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.Let X 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.Let M 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 ≤ ST x hold for all x ∈ M. Similarly, we say that S and T are weakly isotone decreasing if Tx ≥ ST x 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.Let B 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 B r denote the closed ball centered at 0 with radius r.Definition 1.3 see 5 .A function ψ : B X → R is said to be a measure of weak noncompactness if it satisfies the following conditions.
1 The family ker ψ {M ∈ B X : ψ M 0} is nonempty and ker ψ is contained in the set of relatively weakly compact sets of X.
5 If M n n≥1 is a sequence of nonempty weakly closed subsets of X with M 1 bounded and The family ker ψ 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 ker ψ since ψ M ∞ ≤ ψ M n for every n and lim n → ∞ ψ M n 0. Also, it can be easily verified that the measure ψ satisfies where M w is the weak closure of M.
A measure of weak noncompactness ψ is said to be regular if and set additive or have the maximum property if 1.5 The first important example of a measure of weak noncompactness has been defined by De Blasi 6 as follows: Notice that w • is regular, homogeneous, subadditive, and set additive see 6 .
By a measure of noncompactness on a Banach space X 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.Let X 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 ψcondensing.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, every k-ψ-contractive map with k < 1 is ψ-condensing and every ψcondensing map is 1-ψ-contractive.Definition 1.6 see 11 .A map A : D A → X is said to be ws-compact if it is continuous and for any weakly convergent sequence x n n∈N in D A the sequence Ax n 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 map A : D A → X is said to be ww-compact if it is continuous and for any weakly convergent sequence x n n∈N in D A the sequence Ax n n∈N has a weakly convergent subsequence in X. Definition 1.9.Let X 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 Tx − Ty ≤ ϕ x − y 1.7 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 setcontraction 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.Let T be a ww-compact nonlinear contraction mapping on a Banach X.Then for each bounded subset M of X one has Here, w is the De Blasi measure of weak noncompactness.
Proof.Let M be a bounded subset of X and r > w M .There exist 0 ≤ r 0 < r and a weakly compact subset W of X such that M ⊆ W B r 0 .Since T is a φ-nonlinear contraction, then Moreover, since T is ww-compact, then TW w is weakly compact.Accordingly, Letting r → w M and using the continuity of φ we deduce that The following theorem is a sharpening of 3, Theorem 2.1 and 4, Theorem 3.1 .
Theorem 1.12.Let X 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: iii S and T are weakly isotone.
Then T and S have a common fixed point.
Proof.Let x ∈ M be fixed.Consider the sequence {x n } defined by Suppose first that S and T are weakly isotone increasing on M. Then from 1.12 it follows that We show that A 1 is relatively compact.Suppose the contrary, then ψ A 1 > 0. Since T is 1-ψ-contractive and S is ψ-condensing, it follows from 1.14 that

1.15
This is impossible.Thus A 1 is relatively compact.The continuity of S implies that A 2 S A 1 ∪ {x 1 } is relatively compact.Consequently A is relatively compact.Since {x n } 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
To complete the proof we consider the case where S and T are weakly isotone decreasing on M. In this case, the sequence {x n } 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 Theorem 1.12, the modified versions of Theorems 3.6-3.12 in 4 can be proved similarly.
As easy consequences of Theorem 1.12 we obtain the following results.
Corollary 1.15.Let M 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.Let M 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.Let M 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: ii T is a nonexpansive mapping, Abstract and Applied Analysis 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 Corollary 1.17 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 Tx − Ty < x − y whenever x, y ∈ D T with x / y. 1.17 Corollary 1.18.Let M 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: 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 Corollary 1.17 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.Let M be a nonempty closed convex subset of a Banach space X and S, T : X → X be two mappings satisfying the following: 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.Let X 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: iii S and T are weakly isotone.
Then T and S have a common fixed point in M.

1.18
Suppose first that S and T are weakly isotone increasing on M. Then from 1.18 it follows that As in the proof of Theorem 1.12 we set A {x 1 , x 2 , . . ., x n , . ..},A 1 {x 2 , x 4 , . . ., x 2n , . ..}, and We show that A 1 is relatively weakly compact.Suppose the contrary, then ψ A 1 > 0. Since T is 1-ψ-contractive and S is ψ-condensing, it follows from 1.20 that

1.21
This is impossible.Thus A 1 is relatively weakly compact.The weak sequential continuity of S implies that A 2 S A 1 ∪ {x 1 } is relatively weakly compact.The same reasoning as in the proof of Theorem 1.12 gives the desired result.
As easy consequences of Theorem 1.20 we obtain the following results.
Corollary 1.21.Let M 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.Let M 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.Let M 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: 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 Theorem 1.20 on the basis of Lemma 1.11.
Corollary 1.24.Let M 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: 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.Let M 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: 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.Let X 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: 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.Let x ∈ M be fixed.Consider the sequence {x n } defined by Suppose first that S and T are weakly isotone increasing on M. Then from 1.22 it follows that

.}. Clearly
Similar reasoning as in Theorem 1.20 gives that A 1 is relatively weakly compact.From the wscompactness of S it follows that A 2 S A 1 ∪ {x 1 } is relatively compact.Now the continuity of T yields that A 1 T A 2 is relatively compact.Consequently, A A 1 ∪ A 2 is relatively compact.The rest of the proof runs as in the proof of Theorem 1.12.
As easy consequences of Theorem 1.27 we obtain the following results.
Corollary 1.28.Let S, T : X → X be two continuous mappings.Assume T is nonexpansive and wwcompact 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 Theorem 1.27 on the basis of Lemma 1.11.
Corollary 1.29.Let M 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 Theorem 1.27 on the basis of Lemma 1.11.
Corollary 1.30.Let M 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.Let M 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 if X 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.Let X 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: iii S and T are weakly isotone.
Then T and S have a common fixed point in M.
Proof.Let x ∈ M be fixed.Consider the sequence {x n } defined by Suppose first that S and T are weakly isotone increasing on M. Then from 1.25 it follows that Similar reasoning as in Theorem 1.20 gives that A 1 is relatively weakly compact.Thus {x n } has a weakly convergent subsequence, say {x n k }.Since in a Banach lattice every weakly convergent increasing sequence is norm-convergent we infer that {x n k } is norm-convergent.Consequently A 1 is relatively compact.Similar reasoning as in the proof of Theorem 1.12 gives the desired result.

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 p t, x t Integral equations like 2.1 were studied in 20 in L 2 0, 1 and in 4 in L p 0, 1 with 1 < p < ∞.In this section, we look for a nonnegative solution to 2.1 in L 1 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 fixed u ∈ R, the function t → f t, u is measurable from 0, 1 to R, ii for almost any x ∈ 0, 1 , the function f t, • : R → R is continuous.
Let m 0, 1 be the set of all measurable functions x : 0, 1 → R. If f is a Carathéodory function, then f defines a mapping N f : m 0, 1 → m 0, 1 by N f 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.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 1 0 ζ •, s y s ds ∈ L 1 0, 1 whenever y ∈ L 1 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 function p : 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 all r > 0 and for all x ∈ B r we have 0 ≤ θ t a bθ t Note that T and S may be written in the shape: Here N p and N f are the superposition operators associated to the functions p and f, respectively, and K denotes the linear integral defined by Let r 0 be the real defined by r 0 a θ 1 − b θ 2.9 and M be the closed convex subset of L 1 0, 1 defined by Note that, for any x ∈ L 1 0, 1 , the functions Tx and Sx belong to L 1 0, 1 which is a consequence of the assumptions a, b .Moreover, from assumption c it follows that for each x ∈ M we have 0 ≤ p t, x t ≤ x t ≤ r 0 .

2.11
This implies that T M ⊆ M. Also, in view of assumption b we have for all x ∈ M and for all s ∈ 0, 1 Our strategy consists in applying Corollary 1.29 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 N p and N f map continuously L 1 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 ρ n be a weakly convergent sequence of L 1 0, 1 .Using Lemma 2.2, the sequence N f ρ n has a weakly convergent subsequence, say N f ρ n k .Let ρ be the weak limit of N f ρ n k .Accordingly, keeping in mind the boundedness of the mapping ζ t, • we get

2.15
The use of the dominated convergence theorem allows us to conclude that the sequence Sρ n k converges in L 1 0, 1 .
Step 3. S maps bounded sets of L 1 0, 1 into weakly compact sets.To see this, let O be a bounded subset of L 1 0, 1 and let C > 0 such that x ≤ C for all x ∈ O.For x ∈ O we have

2.19
Hence Sx ≥ TSx and Tx ≥ ST x and therefore S and T are weakly isotone decreasing.
Note that the Steps 1-4 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.
Lemma 2.1 see 21, 22 .Let f : 0, 1 ×R → R be a Carathéodory function.Then the superposition operator N f maps continuously L 1 0, 1 into itself if and only if there exist a constant b ≥ 0 and a function a• ∈ L 1 0, 1 such that If f : 0, 1 × R → R is a Carathéodory function and N f maps continuously L 1 0, 1 into itself, then N f is ww-compact.Remark 2.3.Although the Nemytskii operator N f is ww-compact, generally it is not weakly continuous.In fact, only linear functions generate weakly continuous Nemytskii operators in L 1 spaces see, for instance, 23, Theorem 2.6 .
The problem 2.1 may be written in the shape: r {x ∈ L 1 0, 1 : 0 ≤ x t ≤ r for all t ∈ 0, 1 },d b θ ∞ < 1.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 L 1 0, 1 .Proof.
|dt ≤ 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