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Fixedpoint theorems for the sum of two operators under ωcondensing
Fixed Point Theory and Applications volume 2013, Article number: 102 (2013)
Abstract
The purpose of this paper is to establish fixedpoint theorems for the sum of two operators A and B, where the operator A is assumed to be contractive with respect to the measure of weak noncompactness, while B is an φnonlinear contraction. In the last section, we apply such results to study the existence of solutions to a nonlinear Hammerstein integral equation in {L}^{1} space.
1 Introduction
The existence of fixed points for the sum of two operators has been followed with interest for a long time. In 1958, to study the existence of solutions of nonlinear equations of the form
Krasnosel’skii [1] first proved operator A+B has a fixed point whenever M is a nonempty closed convex subset of Banach space X and the operators A and B satisfy:

(i)
A is continuous on M, and A(M) is relatively compact,

(ii)
B is a kcontraction with k\in [0,1),

(iii)
A(M)+B(M)\subset M.
In 1955, Darbo [2] extended the Schauder fixedpoint theorem to the setting of noncompact operators, introducing the notion of kset contraction. It is not hard to see that the Krasnosel’skii theorem is a particular case of the Darbo theorem. Namely, it appears that A+B is a kset contraction with respect to the Kuratowskii measure of noncompactness. In 1967, Sadovskii [3] gave a fixedpoint result more general than the Darbo theorem using the concept of condensing operator.
In 1977, De Blasi [4] introduced the concept of measure of weak noncompactness. Emmanuele [5] established a Sadovskiitype fixedpoint result using the concept of ωcondensing with respect to the measure of weak noncompactness, in which the weak continuity of the operator is required. Recently, GarciaFalset and Latrach established a new version of Sadovskiitype fixedpoint theorem for the weakly sequentially continuous operators (see [[6], Lemma 3.2]).
On the other hand, since the weak continuity condition is usually not easy to verify, Latrach et al. [7, 8] established generalizations of the Schauder, Darbo and Krasnosel’skii fixedpoint theorems for the weak topology. Their analysis uses the concept of the Blasi measure of weak noncompactness. Moreover, and in contrast to previous works, to prove the new versions of the fixedpoint theorems, they neither assume the weak continuity nor the weakly sequential continuity of the operators.
The purpose of this paper is to establish several fixedpoint theorems for the sum of two operators under ωcondensing. By relaxing the condition of weak compactness of operators, these results extend and supplement some previous ones in the literatures.
This paper is organized as follows. In Section 2, we gather some notions and preliminary facts which will be needed in our further considerations. In Section 3, on the basis of a Sadovskiitype fixedpoint theorem for ωcondensing operators and its variant on whole space, we discuss several fixedpoint theorems for the sum of A+B, where A is a ωcontraction and B is an φnonlinear contraction. In Section 4, we apply such results to study the existence of solutions to a nonlinear Hammerstein integral equation in {L}^{1} space.
2 Preliminaries
We first gather together some notations and preliminary facts of some weak topology feature which will be needed in our further considerations. Let \mathcal{B}(X) be the collection of all nonempty bounded subsets of a Banach space X, and let \mathcal{W}(X) be the subset of \mathcal{B}(X) consisting of all weakly compact subsets of X. Also, let {\mathbf{B}}_{r} denote the closed ball in X centered in 0 and with radius r.
De Blasi [4] introduced the map \omega :\mathcal{B}(X)\to {\mathbb{R}}^{+} defined by
Before we launch into the details, we recall some important properties needed hereafter for the sake of completeness (for the proofs, we refer the reader to [4] and [9]).
Lemma 2.1 Let M, {M}_{1} and {M}_{2} be in \mathcal{B}(X); we have:

(a)
\omega \{{M}_{1}\}\le \omega \{{M}_{2}\} whenever {M}_{1}\subset {M}_{2}.

(b)
\omega \{M\}=0 if and only if {\overline{M}}^{w}\in \mathcal{W}(X), where {\overline{M}}^{w} is the weak closure of M.

(c)
\omega \{{\overline{M}}^{w}\}=\omega \{M\}.

(d)
\omega \{co(M)\}=\omega \{M\} where co(M) refers to the convex hull of M.

(e)
\omega \{\lambda M\}=\lambda \omega \{M\}, for all \lambda \in \mathbb{R}.

(f)
\omega \{{M}_{1}+{M}_{2}\}\le \omega \{{M}_{1}\}+\omega \{{M}_{2}\}.

(g)
\omega \{{M}_{1}\cup {M}_{2}\}=max\{\omega \{{M}_{1}\},\omega \{{M}_{2}\}\}.

(h)
If {({M}_{n})}_{n=1}^{\mathrm{\infty}} is a decreasing sequence of nonempty, bounded and weakly closed subsets of X with {lim}_{n\to \mathrm{\infty}}\omega \{{M}_{n}\}=0, then {lim}_{n\to \mathrm{\infty}}{\bigcap}_{k=1}^{\mathrm{\infty}}{M}_{n}\ne \phi and \omega \{{\bigcap}_{k=1}^{n}{M}_{n}\}=0, i.e., {\bigcap}_{k=1}^{\mathrm{\infty}}{M}_{n} is relatively weakly compact.
The map \omega \{\cdot \} is called the De Blasi measure of weak noncompactness. In [9], Appell and De Pascale proved that in {L}^{1}spaces \omega \{\cdot \} has the following form:
for all bounded subsets M of {L}^{1}(\mathrm{\Omega};X), where X is a finite dimensional Banach space and meas(\cdot ) denotes the Lebesgue measure.
Throughout this paper, X denotes a Banach space; \mathcal{D}(T) and \mathcal{R}(T), respectively, denote the domain and range of operator T; and \omega \{S\} denotes the De Blasi measure of weak noncompactness of bounded subset S.
Definition 2.2 An operator T:\mathcal{D}(T)\subset X\to X is said to be ωcontractive (or a ωαcontraction) if it maps bounded sets into bounded sets, and there exists some \alpha \in [0,1) such that \omega \{T(S)\}\le \alpha \omega \{S\} for all bounded sets S in \mathcal{D}(T).
An operator T:\mathcal{D}(T)\subset X\to X is said to be ωcondensing if it maps bounded sets into bounded sets, and \omega \{T(S)\}<\omega \{S\} for all bounded sets S in \mathcal{D}(T) with \omega \{S\}>0.
Remark 2.3 Obviously, every ωαcontraction with 0\le \alpha <1 is ωcondensing.
Let T be an operator from \mathcal{D}(T)\subset X into X. Latrach et al. [7] introduce the following conditions:
(\mathcal{A}1) If {({x}_{n})}_{n\in \mathbb{N}} is a weakly convergent sequence in \mathcal{D}(T), then {(T{x}_{n})}_{n\in \mathbb{N}} has a strongly convergent subsequence in X.
(\mathcal{A}2) If {({x}_{n})}_{n\in \mathbb{N}} is a weakly convergent sequence in \mathcal{D}(T), then {(T{x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence in X.
The conditions (\mathcal{A}1) and (\mathcal{A}2) were already considered in [7, 8, 10–14].
Remark 2.4

(a)
Operators satisfying either (\mathcal{A}1) or (\mathcal{A}2) are not necessarily weakly continuous.

(b)
An operator satisfies (\mathcal{A}1) if and only if it maps relatively weakly compact sets into relatively strongly compact ones.

(c)
An operator satisfies (\mathcal{A}2) if and only if it maps relatively weakly compact sets into relatively weakly compact ones (EberleinŠmulian theorem, see, e.g., [[15], pp.248250]).

(d)
Every ωcontractive operator satisfies (\mathcal{A}2).

(e)
The condition (\mathcal{A}2) holds true for every bounded linear operator.
Lemma 2.5 Let operator T:\mathcal{D}(T)\subset X\to X satisfy (\mathcal{A}1), and let operator Q:\mathcal{R}(T)\subset X\to X be continuous. Then the compound operator Q\circ T:\mathcal{D}(T)\subset X\to X satisfies (\mathcal{A}1).
Proof Let {({x}_{n})}_{n\in \mathbb{N}} be a weakly convergent sequence in \mathcal{D}(T). By the hypothesis of T satisfying (\mathcal{A}1), {(T{x}_{n})}_{n\in \mathbb{N}} has a strongly convergent subsequence, say {(T{x}_{{n}_{k}})}_{k\in \mathbb{N}}. The continuity of Q implies that {(QT{x}_{{n}_{k}})}_{k\in \mathbb{N}} is also strongly convergent, and therefore Q\circ T satisfies (\mathcal{A}1). □
Definition 2.6 An operator T:\mathcal{D}(T)\subset X\to X is said to be φnonlinear contractive (or an φnonlinear contraction), if there exists a continuous and nondecreasing function \phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} such that
where \phi (r)<r for r>0.
Remark 2.7 Obviously, all strictly contractions are φnonlinear contractions.
Lemma 2.8 (see [[16], Lemma 1.11])
Let operator T:\mathcal{D}(T)\subset X\to X be φnonlinear contractive on a Banach space X and satisfy (\mathcal{A}2). Then for each bounded subset M of X one has \omega \{T(M)\}\le \phi (\omega \{M\}).
Lemma 2.9 If an operator T:X\to X is φnonlinear contractive, then F:=IT is a homeomorphism of X onto X.
Proof For any y\in X, we define the operator {\tau}_{y} from X to X, by {\tau}_{y}x=Tx+y. Since T is φnonlinear contractive, it is easy to see that {\tau}_{y} is also φnonlinear contractive. According to Theorem 1 in [17], {\tau}_{y} has a unique fixed point x such that x=Tx+y, i.e. y=(IT)x, and then F:=IT is surjective on X.
If x,y\in X and x\ne y, then
which implies that F is injective and {F}^{1} exists on X.
For proving the continuity of {F}^{1}, suppose that there exists a point x and a sequence {({x}_{n})}_{n\in \mathbb{N}} in X such that F{x}_{n}\to Fx, and {lim}_{n\to \mathrm{\infty}}{sup}_{k\ge n}\parallel {x}_{k}x\parallel =a. Consequently, from the inequality
we obtain that 0\ge a\phi (a), which implies that a=0 and, therefore, {F}^{1} is continuous. □
3 Fixedpoint theorems for the sum of two operators
The following theorem was proved by Ben Amar and GarciaFalset [13], and its more general form was presented by Agarwal et al. [14] is a variant of the Sadovskii fixedpoint theorem for the classes of operators which satisfy (\mathcal{A}1).
Theorem 3.1 (see [[13], Theorem 3.1] or [[14], Theorem 2.1 and Corollary 2.2])
Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Assume that T:M\to M is continuous and satisfies (\mathcal{A}1). If T is ωcondensing, then it has a fixed point in M.
Our purpose here is to establish a fixedpoint theorem for the sum of a ωcontractive operator and an φnonlinear contractive operator.
Theorem 3.2 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction with \phi (r)<(1\alpha )r for r>0, and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Proof By Lemma 2.9, IB has a continuous inverse defined on X, and then J:={(IB)}^{1}A is well defined on M. Once we prove that J has a fixed point in M, the proof is achieved.
For any y\in M, according to Lemma 2.9 there exists a unique x\in X such that Ay=xBx. The hypothesis (iii) shows that x\in M, which implies that A(M)\subset (IB)(M) and, therefore, J(M)\subset M.
Obviously, the compound operator J is continuous since A and {(IB)}^{1} is continuous and by Lemma 2.5, J satisfies (\mathcal{A}1). Now by referring to the formula
for every subset S of M with \omega \{S\}>0, we have
Since A is ωαcontractive and B satisfies \omega \{BJ(S)\}\le \phi (\omega \{J(S)\}) by Lemma 2.8, we have
Now, if \alpha =0, inequality (3.1) becomes \omega \{J(S)\}\le \phi (\omega \{J(S)\}), which implies that \omega \{J(S)\}=0. Otherwise, by recalling the assumption that \phi (r)<(1\alpha )r for r>0, inequality (3.1) becomes
In both cases, J is shown to be ωcondensing. Now the use of Theorem 3.1 achieves the proof. □
Remark 3.3 It should be noticed to the following particular cases:

(1)
If we take B=0, then we return the above theorem back to [[7], Theorem 2.2], which is an extension of the Darbo fixedpoint theorem for ωcontractive operators.

(2)
If we take \alpha =0 and the function \phi (r)=\beta r (0\le \beta <1) in the above theorem, we obtain a result which was [[7], Theorem 2.3].

(3)
If we only take the function \phi (r)=\beta r (0\le \beta <1\alpha) in the above theorem, we obtain the following Corollary 3.4, which is a new fixedpoint theorem for the sum of two operators.

(4)
If we only take \alpha =0 in the above theorem, we obtain the following Corollary 3.6, which is the new version of Krasnosel’skiitype fixedpoint theorems.
Corollary 3.4 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is a strict contraction with \beta \in [0,1\alpha ), and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.5 The above corollary is a variant and supplement of [[6], Theorem 3.3], in which the authors demand that the operators A and B are weakly sequentially continuous.
Corollary 3.6 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is a continuous, A(M) is relatively weakly compact and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction, and B satisfies (\mathcal{A}2),

(iii)
(x=Bx+Ay,y\in M)\Rightarrow x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.7 The above corollary is a variant and supplement of [[18], Theorem 2.1], in which the author demands that the operators A and B are weakly sequentially continuous, and B is strictly contractive.
Now, on the basis of Corollary 3.6, we prove the following fixedpoint theorem for the sum of a weaklystrongly continuous operator and a nonexpansive operator.
Theorem 3.8 Let M be a nonempty, bounded, closed and convex subset of a Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is weaklystrongly continuous, and A(M) is relatively weakly compact,

(ii)
B is nonexpansive and ωcondensing,

(iii)
IB is demiclosed,

(iv)
if \lambda \in (0,1) and x=\lambda Bx+Ay for some y\in M, then x\in M.
Then there is a point x\in M such that Ax+Bx=x.
Remark 3.9

(1)
Recall that an operator T:\mathcal{D}(T)\subset X\to X is said to be demiclosed if for any sequence {({x}_{n})}_{n\in \mathbb{N}} in D(T) that {x}_{n}\rightharpoonup x and T{x}_{n}\to y, then x\in \mathcal{D}(T) and Tx=y.

(2)
The assumption (iv) in the above theorem was first introduced in [19], it is slight different with [[10], Corollary 3.1] and [[11], Theorem 2.1].

(3)
In [[11], Theorem 2.1] and [[18], Theorem 2.4], the following condition is required: if {({x}_{n})}_{n\in \mathbb{N}} is a sequence of M such that {({x}_{n}T{x}_{n})}_{n\in \mathbb{N}} is weakly convergent, then {({x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence. In the above theorem, we replaced it with the ωcondensing of B.
Proof of Theorem 3.8 For each \lambda \in (0,1), the operators A and λB fulfill the conditions of Corollary 3.6 and, therefore, there is a point {x}_{\lambda}\in M such that {x}_{\lambda}=\lambda B{x}_{\lambda}+A{x}_{\lambda}. Now choose a sequence {({\lambda}_{n})}_{n\in \mathbb{N}}\subset (0,1) such that {\lambda}_{n}\to 1. Consequently, there exists a sequence {({x}_{n})}_{n\in \mathbb{N}}\subset M such that
Let S=\{{x}_{n}:n\in \mathbb{N}\}. We claim that S is relatively weakly compact. Suppose that it is not the case, by assumption (i) and (ii), we have
This contradiction tells us that the sequence {({x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence, i.e., there exists {({x}_{{n}_{k}})}_{k\in \mathbb{N}} such that {x}_{{n}_{k}}\rightharpoonup x\in M. By assumption (i), we have A{x}_{{n}_{k}}\to Ax, and then (I{\lambda}_{{n}_{k}}B){x}_{{n}_{k}}\to Ax. Since {({x}_{{n}_{k}})}_{k\in \mathbb{N}} is contained in bounded set M, and B maps M into a bounded set (B is nonexpansive), then \parallel B{x}_{{n}_{k}}\parallel is norm bounded. Thus, we have (1{\lambda}_{{n}_{k}})\parallel B{x}_{{n}_{k}}\parallel \to 0. Moreover, we have
that is, (IB){x}_{{n}_{k}}\to Ax. By assumption (iii), we have (IB)x=Ax, and then the proof is achieved. □
If the Banach space X is reflexive, then B is always ωcondensing on M (see, e.g., [[15], p.251]). Moreover, if we supposed that X is uniformly convex Banach space, then IB:M\to X is demiclosed (see, e.g., [[20], pp.476478]). Thus, we obtain the following consequence.
Corollary 3.10 Let M be a nonempty, bounded, closed and convex subset of a uniformly convex Banach space X. Suppose that A:M\to X and B:X\to X are two operators such that

(i)
A is weaklystrongly continuous,

(ii)
B is nonexpansive,

(iii)
if \lambda \in (0,1) and x=\lambda Bx+Ay for some y\in M, then x\in M.
Then there is a point x\in M such that Ax+Bx=x.
In order to use the above results on the whole space, we first prove the following result.
Theorem 3.11 Let X be a Banach space X. Assume that the operator T:X\to X be continuous ωcondensing and satisfies (\mathcal{A}1). Then either

(a)
equation x=Tx has a solution, or

(b)
the set \{x\in X:x=\lambda T(x)\} is unbounded for some \lambda \in (0,1).
Proof Choose an arbitrary R>0. Define for each x\in X
Clearly, ρ is a continuous retraction of X on {\mathbf{B}}_{R}. Thus, we can define the mapping {T}_{\rho}:{\mathbf{B}}_{R}\to {\mathbf{B}}_{R} by {T}_{\rho}x=\rho (Tx).
Since T and ρ are continuous, obviously {T}_{\rho} is also continuous. Furthermore, since T satisfies (\mathcal{A}1) and ρ is continuous, hence {T}_{\rho} also satisfies (\mathcal{A}1). We next claim that \omega \{{T}_{\rho}(S)\}<\omega \{S\} for any S\subset {\mathbf{B}}_{r} with \omega \{S\}>0.
Indeed, {T}_{\rho}({\mathbf{B}}_{r})=\rho (T({\mathbf{B}}_{r})). For any x\in {\mathbf{B}}_{r}, there are two possibilities:

(1)
\parallel Tx\parallel \le R; in this case, \rho (Tx)=Tx\in T(S)\subset co(T(S)\cup \{\mathbf{0}\}).

(2)
\parallel Tx\parallel >R; in this case, \rho (Tx)=\frac{R}{\parallel Tx\parallel}Tx=\frac{R}{\parallel Tx\parallel}Tx+(1\frac{R}{\parallel Tx\parallel})\cdot \mathbf{0}\in co(T(S)\cup \{\mathbf{0}\}).
The above argument yields {T}_{\rho}(S)\subset co(T(S)\cup \{\mathbf{0}\}).
Now, by using the properties of the measure of weak noncompactness and properties of T, we have that
as claimed, that is, {T}_{\rho} is a ωcondensing.
The above argument shows that {T}_{\rho}:{\mathbf{B}}_{r}\to {\mathbf{B}}_{r} is under the conditions of Theorem 3.1 and thus we have that there exists {x}_{0}\in {\mathbf{B}}_{r} such that {x}_{0}={T}_{\rho}{x}_{0}. Indeed, we obtain the following results:

(a)
if T{x}_{0}\in {\mathbf{B}}_{R}, then {x}_{0}={T}_{\rho}{x}_{0}=\rho (T{x}_{0})=T{x}_{0}, that is, T has a fixed point; otherwise,

(b)
if T{x}_{0}\notin {\mathbf{B}}_{R}, then {x}_{0}={T}_{\rho}{x}_{0}=\rho (T{x}_{0})=\frac{R}{\parallel T{x}_{0}\parallel}T{x}_{0}, that is, {x}_{0} is a solution of the equation x=\lambda Tx for \lambda =\frac{R}{\parallel T{x}_{0}\parallel}\in (0,1) and \parallel {x}_{0}\parallel =R.
Consequently, if there is no x\in {\mathbf{B}}_{R} such that x=Tx for any R>0, then the above arguments show that the set of solutions of equation x=\lambda T(x) is unbounded for some \lambda \in (0,1). □
We are now in a position to prove the main result on the whole space.
Theorem 3.12 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction with \phi (r)<(1\alpha )r for r>0, and B satisfies (\mathcal{A}2),

(iii)
function φ satisfies {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty}.
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).
Remark 3.13 Obviously, assumption (iii) in the above theorem is unnecessary whenever \alpha \ne 0.
Proof of Theorem 3.12 As in the proof of Theorem 3.2, it can be seen that the compound operator J:={(IB)}^{1}A is well defined from X into X. Clearly, J is continuous and J satisfies (\mathcal{A}1) by Lemma 2.5.
Let us prove that J maps bounded set into a bounded set. For any bounded set S such that u,v\in {(IB)}^{1}A(S), there exist x,y\in S such that u={(IB)}^{1}Ax and v={(IB)}^{1}Ay, that is, uBu=Ax and vBv=Ay. Thus, by assumption (ii) and the boundness of A(S), we have
Suppose that J(S)={(IB)}^{1}A(S) is unbounded, i.e., there exist sequences {({u}_{n})}_{n\in \mathbb{N}} and {({v}_{n})}_{n\in \mathbb{N}} such that \parallel {u}_{n}{v}_{n}\parallel \to +\mathrm{\infty}, and then \parallel {u}_{n}{v}_{n}\parallel \phi (\parallel {u}_{n}{v}_{n}\parallel )\to +\mathrm{\infty} by assumption (iii). This is a contradiction with diam(A(S))<+\mathrm{\infty} in (3.2) and, therefore, J(S) is bounded.
It is similar to that of Theorem 3.2 to prove that \omega \{J(S)\}<\omega \{S\} for every bounded set S in \mathcal{D}(T) with \omega \{S\}>0. Now, by using Theorem 3.11 for operator J, we obtain that either

(a)
the equation x=Ax+Bx has a solution which is the solution of equation x=Jx, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\}=\{x\in X:x=\lambda {(IB)}^{1}Ax\} is unbounded for some \lambda \in (0,1).
□
Corollary 3.14 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous ωαcontraction with \alpha \in [0,1), and A satisfies (\mathcal{A}1),

(ii)
B is a strict contraction with \beta \in [0,1\alpha ), and B satisfies (\mathcal{A}2).
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).
Corollary 3.15 Let X be a Banach space. Suppose that A,B:X\to X are two operators such that

(i)
A is a continuous, A(M) is relatively weakly compact, and A satisfies (\mathcal{A}1),

(ii)
B is an φnonlinear contraction, and B satisfies (\mathcal{A}2),

(iii)
function φ satisfies {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty}.
Then, either

(a)
the equation x=Ax+Bx has a solution, or

(b)
the set \{x\in X:x=\lambda B(x/\lambda )+\lambda A(x)\} is unbounded for some \lambda \in (0,1).
4 Application to Hammerstein integral equations in {L}^{1} space
Let Ω be a domain of {\mathbb{R}}^{n}. A function f:\mathrm{\Omega}\times X\to Y is said to be a Carathéodory function if

(i)
for any fixed x\in X, the function t\to f(t,x) is measurable from Ω to Y;

(ii)
for almost any t\in \mathrm{\Omega}, the function f(t,\cdot ):X\to Y is continuous.
Let m(\mathrm{\Omega}) be the set of all measurable functions \psi :\mathrm{\Omega}\to X. If f is a Carathéodory function, then f defines an operator {\mathbf{N}}_{f}:m(\mathrm{\Omega})\to m(\mathrm{\Omega}) by {\mathbf{N}}_{f}(\psi )(t)\to f(t,\psi (t)). This operator is called the Nemytskii operator associated to f (or the superposition operator). Regarding its continuity and weak compactness, we have the following lemma.
Lemma 4.1 Let X, Y be two finite dimensional Banach spaces. If f:\mathrm{\Omega}\times X\to Y is a Carathéodory function, then the Nemytskii operator {\mathbf{N}}_{f} maps {L}^{1}(\mathrm{\Omega};X) into {L}^{1}(\mathrm{\Omega};Y) if and only if there exist a constant b\ge 0 and a function a\in {L}_{+}^{1}(\mathrm{\Omega}) such that
where {L}_{+}^{1}(\mathrm{\Omega}) denotes the positive cone of the space {L}^{1}(\mathrm{\Omega}) (see [21]or [22]).
With the conditions of Lemma 4.1, the operator {\mathbf{N}}_{f} is obviously continuous and maps bounded sets of {L}^{1}(\mathrm{\Omega};X) into bounded sets of {L}^{1}(\mathrm{\Omega};Y).
Lemma 4.2 (see [[8], Lemma 3.2])
Let Ω be a bounded domain in {\mathbb{R}}^{n}. If f:\mathrm{\Omega}\times X\to Y is a Carathéodory function and {\mathbf{N}}_{f} maps {L}^{1}(\mathrm{\Omega};X) into {L}^{1}(\mathrm{\Omega};Y), then {\mathbf{N}}_{f} satisfies (\mathcal{A}2).
Remark 4.3 Although Nemytskii operator {\mathbf{N}}_{f} satisfies (\mathcal{A}2), generally it is not weakly continuous. In fact, only linear functions generate weakly continuous Nemytskii operators in {L}^{1} spaces (see, for instance, [[9], Theorem 2.6]). The question of considering the weak sequential continuity of the Nemytskii operator acting from space {L}^{p} to space {L}^{q} (1\le p,q<\mathrm{\infty}) is discussed in [22] and the answer is shown to be negative at least for p=2.
Next, we give an example of application for Theorem 3.12 in the Banach space of integrable function {L}^{1}(\mathrm{\Omega};X).
Example 4.4 We will study now the existence of solutions for the following variant of Hammerstein’s integral equation
in {L}^{1}(\mathrm{\Omega};X), the space of Lebesgue integrable functions on a measurable subset Ω of {\mathbb{R}}^{n} with values in a finite dimensional Banach space X. Here, f is a nonlinear function and k is measurable, while g is a function satisfying φnonlinear contractive condition in {L}^{1}(\mathrm{\Omega};X).
First, observe that the above problem may be written in the form
where B is the Nemytskii operator associated to the function g (i.e., B={\mathbf{N}}_{g}) from {L}^{1}(\mathrm{\Omega};X) into {L}^{1}(\mathrm{\Omega};X) by
and A=\mu L{\mathbf{N}}_{f} is the product of the Nemytskii operator associated to f and the linear integral operator μL where \mu \in \mathbb{C} and L is defined from {L}^{1}(\mathrm{\Omega};Y) into {L}^{1}(\mathrm{\Omega};X) by
Let us now introduce the following assumptions.
Assumptions 4.5

(a)
f is a Carathéodory function and {\mathbf{N}}_{f} acts from {L}^{1}(\mathrm{\Omega};X) into {L}^{1}(\mathrm{\Omega};Y);

(b)
b\mu \parallel L\parallel <1 (the constant b was introduced in Lemma 4.1);

(c)
g:\mathrm{\Omega}\times X\to X is a measurable function with g(\cdot ,0)\in {L}^{1}(\mathrm{\Omega};X), and there exists a continuous and nondecreasing function \phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} such that
{\int}_{\mathrm{\Omega}}{\parallel g(t,x)g(t,y)\parallel}_{X}\phantom{\rule{0.2em}{0ex}}dt\le \phi \left({\parallel xy\parallel}_{{L}^{1}(\mathrm{\Omega};X)}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}x,y\in X,
where \phi (r)<(1b\mu \parallel L\parallel )r for r>0 and {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty};

(d)
the function k:\mathrm{\Omega}\times \mathrm{\Omega}\to L(Y,X) is strongly measurable and the linear operator L defined by (4.2) maps {L}^{1}(\mathrm{\Omega};Y) into {L}^{1}(\mathrm{\Omega};X);

(e)
the functions \rho (t):\mathrm{\Omega}\to L(Y,X), s\to \rho (t)(s):=k(t,s) belong to {L}^{\mathrm{\infty}}(\mathrm{\Omega};L(Y,X));

(f)
the solution of the integral equation
\psi (t)=\lambda g(t,\frac{1}{\lambda}\psi (t))+\lambda \mu {\int}_{\mathrm{\Omega}}k(t,s)f(s,\psi (s))\phantom{\rule{0.2em}{0ex}}ds
is bounded for \lambda \in (0,1).
Remark 4.6 (1) Clearly, the condition that {lim}_{r\to +\mathrm{\infty}}[r\phi (r)]=+\mathrm{\infty} in assumption (c) is unnecessary whenever b\mu \parallel L\parallel \ne 0;
(2) It should be noted that assumptions (d) and (e) lead to the estimate
and so
This shows that the linear operator L is continuous, hence weakly continuous from {L}^{1}(\mathrm{\Omega};Y) into {L}^{1}(\mathrm{\Omega};X) and that \parallel L\parallel \le {\parallel \rho \parallel}_{{L}^{1}(\mathrm{\Omega};{L}^{\mathrm{\infty}})}.
(3) By assumption (c), we get
for every u\in X. This shows that the Nemytskii operator {\mathbf{N}}_{g} is continuous and maps bounded sets of {L}^{1}(\mathrm{\Omega};X) into bounded sets of {L}^{1}(\mathrm{\Omega};X). According to Lemma 4.2, the operator B satisfies (\mathcal{A}2).
Now we are in a position to state our main result.
Theorem 4.7 Let X and Y be two finite dimensional Banach spaces and Ω be a bounded domain of {\mathbb{R}}^{n}. Assume that the conditions (a)(f) are satisfied, then the problem (4.1) has at least one solution in {L}^{1}(\mathrm{\Omega};X).
Proof Let us first observe that Lemma 4.1 implies that there are a\in {L}_{+}^{1}(\mathrm{\Omega}) and b>0 such that
So, for any bounded subset S of {L}^{1}(\mathrm{\Omega};X), we have
According to (2.1), this leads \omega \{{\mathbf{N}}_{f}(S)\}\le b\omega \{S\}. Thus, we have
which implies that A is a ωb\mu \parallel L\parallelcontractive.
On the other hand, clearly A is continuous (see Lemma 4.1 and Remark 4.6(2)). Now we check that A satisfies the condition (\mathcal{A}1). For this end, let {({x}_{n})}_{n\in \mathbb{N}} be a weakly convergent sequence of {L}^{1}(\mathrm{\Omega};X). Using the fact that {\mathbf{N}}_{f} satisfies (\mathcal{A}2), and then {({\mathbf{N}}_{f}{x}_{n})}_{n\in \mathbb{N}} has a weakly convergent subsequence, say {({\mathbf{N}}_{f}{x}_{{n}_{k}})}_{k\in \mathbb{N}}. Moreover, the continuity of the linear operator L implies its weak continuity on {L}^{1}(\mathrm{\Omega};Y). Thus, the sequence {(L{\mathbf{N}}_{f}{x}_{{n}_{k}})}_{k\in \mathbb{N}}, i.e. {(A{x}_{{n}_{k}})}_{k\in \mathbb{N}} converges pointwisely for a.e. t\in \mathrm{\Omega}. Using Vitali’s convergence theorem, we conclude that {(A{x}_{{n}_{k}})}_{k\in \mathbb{N}} converges strongly in {L}^{1}(\mathrm{\Omega};X). Therefore, A satisfies (\mathcal{A}1).
Let x(t),y(t)\in {L}^{1}(\mathrm{\Omega};X). It follows from assumption (c) that
So, B is φnonlinear contractive on {L}^{1}(\mathrm{\Omega};X) and from Remark 4.6(3), B satisfies (\mathcal{A}2).
The above arguments show that A and B satisfy the conditions of Theorem 3.12, and assumption (f) allows us to affirm that equation (4.1) has a solution. □
Remark 4.8 Equation (4.1) was respectively considered in [8] and [12] under different assumptions. However, our results relax some assumptions of [[8], Theorem 3.1] and [[12], Theorem 3.1].
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Wang, F. Fixedpoint theorems for the sum of two operators under ωcondensing. Fixed Point Theory Appl 2013, 102 (2013). https://doi.org/10.1186/168718122013102
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DOI: https://doi.org/10.1186/168718122013102
Keywords
 measure of weak noncompactness
 weak condensing
 φnonlinear contraction
 nonlinear Hammerstein integral equations