# Krasnosel’skii-type fixed point theorems with applications to Volterra integral equations

- Nawab Hussain
^{1}Email author and - Mohamed Aziz Taoudi
^{2, 3}

**2013**:196

https://doi.org/10.1186/1687-1812-2013-196

© Hussain and Taoudi; licensee Springer 2013

**Received: **24 April 2013

**Accepted: **5 July 2013

**Published: **22 July 2013

## Abstract

In this paper we present some fixed point results for the sum $S+T$ of two mappings where *S* is a strict contraction and *T* is not necessarily weakly compact and satisfies a new condition formulated in terms of an axiomatic measure of weak noncompactness. Our fixed point results extend and improve several earlier results in the literature. In particular, our results encompass the analogues of Krasnosel’skii’s and Sadovskii’s fixed point theorems for sequentially weakly continuous mappings and a number of their generalizations. Finally, an application to integral equations is given to illustrate the usability of the obtained results.

**MSC:**37C25, 40D05, 31B10.

## Keywords

*w*-contractivecondensing mappingfixed point

## Dedication

Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday

## 1 Introduction

*S*is a strict contraction while

*T*is power-convex condensing w.r.t.

*S*. In some applications, the concepts of ws-compactness and ww-compactness seem to be more practical than the sequential weak continuity. These concepts arise naturally in the study of both integral and partial differential equations in nonreflexive Banach spaces (see [7, 14–16] and the references therein). We prove some fixed point theorems for the sum $S+T$ where

*S*is a ww-compact strict contraction while

*T*is ws-compact and power-convex condensing w.r.t.

*S*without the sequential weak continuity of the involved mappings. The presented results extend all the fixed point theorems quoted above. As an application, we investigate the existence of continuous solutions to a perturbed Volterra integral equation which extends the corresponding results of [13] and many others. For convenience, we first recall some basic concepts and notations. Let

*E*be a Banach space, let ${\mathrm{\Omega}}_{E}$ be the collection of all nonempty bounded subsets of

*E*, and let ${\mathcal{W}}_{E}$ be the subset of ${\mathrm{\Omega}}_{E}$ consisting of all weakly compact subsets of

*E*. Let ${B}_{r}$ denote the closed ball in

*E*centered at 0 with radius $r>0$. The De Blasi [3] measure of weak noncompactness is the map $w:{\mathrm{\Omega}}_{E}\to [0,\mathrm{\infty})$ defined by

*w*which we will tacitly use in the sequel. Let $A,B\in {\mathrm{\Omega}}_{E}$, then we have the following:

- (1)
$w(A)=0$ if and only if

*A*is relatively weakly compact; - (2)
$w(A)=w({\overline{A}}^{w})$, where ${\overline{A}}^{w}$ is the weak closure of

*A*; - (3)
$w(\mathit{co}(A))=w(A)$, where $\mathit{co}(A)$ denotes the convex hull of

*A*; - (4)
if $A\subset B$, then $w(A)\le w(B)$;

- (5)
$w(A\cup B)=max\{w(A),w(B)\}$;

- (6)
$w(\lambda A)=|\lambda |w(A)$ for $\lambda \in \mathbb{R}$, where $\lambda A=\{\lambda x:x\in A\}$;

- (7)
$w(A+B)\le w(A)+w(B)$, where $A+B=\{x+y:x\in A,y\in B\}$;

- (8)
if ${A}_{n}$ is a sequence of nonempty, weakly closed subsets of

*E*with ${A}_{1}$ bounded and ${A}_{1}\supseteq {A}_{2}\supseteq \cdots \supseteq {A}_{n}\cdots $ with ${lim}_{n\to \mathrm{\infty}}w({A}_{n})=0$, then ${\bigcap}_{n=1}^{\mathrm{\infty}}{A}_{n}\ne \mathrm{\varnothing}$ and $w({\bigcap}_{n=1}^{\mathrm{\infty}}{A}_{n})=0$.

*E*be a Banach space, let

*M*be a nonempty closed convex subset of

*E*, and let $S,T:M\to E$ be two nonlinear mappings and ${x}_{0}\in E$. For any $N\subseteq M$, we set

for $n=2,3,\dots $ .

**Definition 1.1**Let

*E*be a Banach space, let

*M*be a nonempty closed convex subset of

*E*, and let

*μ*be a measure of weak noncompactness on

*E*. Let $T,S:M\to E$ be two bounded mappings (

*i.e.*, they take bounded sets into bounded ones) and ${x}_{0}\in M$. We say that

*T*is an

*S*-convex-power condensing operator about ${x}_{0}$ and ${n}_{0}$ w.r.t.

*μ*if for any bounded set $N\subseteq M$ with $\mu (N)>0$ we have

Obviously, $T:M\to M$ is power-convex condensing with respect to *μ* about ${x}_{0}$ and ${n}_{0}$ [13] if and only if it is a 0-convex-power condensing operator about ${x}_{0}$ and ${n}_{0}$ w.r.t. *μ*.

The following results are crucial for our purposes. We first state a theorem of Ambrosetti type (see [17] for a proof).

**Theorem 1.1**

*Let*

*E*

*be a Banach space and let*$H\subseteq C([0,T],E)$

*be bounded and equicontinuous*.

*Then the map*$t\to w(H(t))$

*is continuous on*$[0,T]$

*and*

*where* $H(t)=\{h(t):h\in H\}$ *and* $H[0,T]={\bigcup}_{t\in [0,T]}\{h(t):h\in H\}$.

**Theorem 1.2** [[18], Theorem 9]

*Let* *S* *be a Hausdorff compact space and* *E* *be a Banach space*. *A bounded sequence* $({f}_{n})\subset C(S,E)$ *converges weakly to* $f\in C(S,E)$ *if and only if*, *for every* $t\in S$, *the sequence* $({f}_{n}(t))$ *converges weakly* (*in* *E*) *to* $f(t)$.

**Definition 1.2** A mapping $T:D(T)\subset X\to X$ is called *k-Lipschitzian* if $\parallel Tx-Ty\parallel \le k\parallel x-y\parallel $ for all $x,y\in D(T)$. *T* is called *strict contraction* if $k\in [0,1)$ and *nonexpansive* if $k=1$.

**Lemma 1.1** [11]

*Let* *M* *be a subset of* *E* *and let* $T:M\to E$ *be a* *k*-*Lipschitzian map*. *Assume that* *T* *is a sequentially weakly continuous map*. *Then* $w(T(A))\le kw(A)$ *for each bounded subset* *A* *of* *M*; *here*, $w(\cdot )$ *stands for the De Blasi measure of weak noncompactness*.

**Definition 1.3** We say that $T:D(T)\subset X\to X$ is demiclosed if for any sequence $\{{x}_{n}\}$ weakly convergent to an element ${x}^{\ast}\in D(T)$ with $\{T{x}_{n}\}$ norm-convergent to an element *y*, then $T{x}^{\ast}=y$.

**Theorem 1.3** [[19], Theorem 5.1.2]

*Let* *M* *be a bounded closed convex subset of a Banach space* *X*, *and let* *T* *be a nonexpansive mapping of* *M* *into* *M*. *Then*, *for each* $\epsilon >0$, *there is an* ${x}_{\epsilon}\in M$ *such that* $\parallel T{x}_{\epsilon}-{x}_{\epsilon}\parallel <\epsilon $.

## 2 Fixed point theory for the sum in the weak topology

**Theorem 2.1**

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*,

*and let*

*μ*

*be a measure of weak noncompactness on*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two mappings satisfying*:

- (i)
*T**is sequentially weakly continuous*, - (ii)
*S**is a strict contraction*, - (iii)
*there are an integer*${n}_{0}$*and a vector*${x}_{0}\in E$*such that**T**is**S*-*power*-*convex condensing w*.*r*.*t*.*μ**about*${x}_{0}$*and*${n}_{0}$, - (iv)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$, - (v)
*if*$\{{x}_{n}\}$*is a sequence in*${\mathcal{F}}^{({n}_{0},{x}_{0})}(T,S,M)$*such that*${x}_{n}\rightharpoonup x$,*then*$S{x}_{n}\rightharpoonup Sx$.

*Then* $T+S$ *has at least one fixed point in* *M*.

*Proof*Let $y\in M$. The map which assigns to each $x\in E$ the value $Sx+Ty$ defines a strict contraction mapping from

*E*into itself and so it has a unique fixed point in

*E*by the contraction mapping principle [19]. Let us denote by $\tau :M\to E$ the map which assigns to each $y\in M$ the unique point in

*M*such that $\tau (y)=S\tau (y)+Ty$. From assumption (iv) we infer that $\tau (M)\subset M$. Notice

*n*we have

*C*is a closed convex subset of

*M*and $\tau (C)\subseteq C$. Thus $C\in \mathrm{\Gamma}$. This implies $\overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\})\subseteq C$. Hence $\tau (\overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\}))\subseteq \tau (C)\subseteq \overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\})$. Consequently, $\overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\})\in \mathrm{\Gamma}$. Hence $C\subseteq \overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\})$. As a result $\overline{\mathit{co}}(\tau (C)\cup \{{x}_{0}\})=C$. This shows that $\mathcal{P}(1)$ holds. Let

*n*be fixed and suppose $\mathcal{P}(n)$ holds. This implies

which yields that *C* is weakly compact. We claim now that $\tau :C\to C$ is sequentially weakly continuous. Indeed, let $\{{y}_{n}\}$ be a sequence in *C* such that ${y}_{n}\rightharpoonup y$ in *C*. Since $\tau (C)\subseteq C$, then there exists a subsequence $\{{y}_{{n}_{k}}\}$ of $\{{y}_{n}\}$ such that $\tau {y}_{{n}_{k}}\rightharpoonup z$ for some $z\in C$. By (v) $S\tau {y}_{{n}_{k}}\rightharpoonup Sz$. Also from (i) it follows that $T{y}_{{n}_{k}}\rightharpoonup Ty$ and hence the equality $\tau {y}_{{n}_{k}}=S\tau ({y}_{{n}_{k}})+T{y}_{{n}_{k}}$ gives us $z=Sz+Ty$. By uniqueness, we conclude that $\tau (y)=z$.

*y*. Then, arguing as before, we may extract a subsequence $({y}_{{n}_{{j}_{k}}})$ of $({y}_{{n}_{j}})$ such that $\tau ({y}_{{n}_{{j}_{k}}})\rightharpoonup \tau (y)$, which is absurd since $\tau ({y}_{{n}_{{j}_{k}}})\notin {N}^{w}$ for all $k\ge 1$. Finally,

*τ*is weakly sequentially continuous. Applying the Arino-Gautier-Penot fixed point theorem [5], we infer that there exists $x\in C$ such that

□

It is worthwhile to emphasize that Theorem 2.1 encompasses a lot of previously known results. In particular, if we take $B=0$ in Theorem 2.1, we recapture the following fixed point theorem, which was proved in [[13], Theorem 2.1].

**Corollary 2.1** *Let* *M* *be a nonempty bounded closed convex subset of a Banach space* *X*. *Suppose that* $T:M\to M$ *is weakly sequentially continuous and there exist an integer* ${n}_{0}$ *and a vector* ${x}_{0}\in E$ *such that* *T* *is power*-*convex condensing about* ${x}_{0}$ *and* ${n}_{0}$. *Then* *T* *has at least one fixed point in* *M*.

Another consequence of Theorem 2.1 is the following result, which is a sharpening of [[20], Theorem 2.9].

**Corollary 2.2**

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two mappings satisfying*:

- (i)
*T**is sequentially weakly continuous*, - (ii)
*S**is a strict contraction with constant**k*, - (iii)
*there exists an integer*${n}_{0}$*such that*${\mathcal{F}}^{({n}_{0},{x}_{0})}(T,S,M)$*is relatively weakly compact*, - (iv)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$, - (v)
*if*$\{{x}_{n}\}$*is a sequence in*${\mathcal{F}}^{({n}_{0},{x}_{0})}(T,S,M)$*such that*${x}_{n}\rightharpoonup x$,*then*$S{x}_{n}\rightharpoonup Sx$.

*Then* $T+S$ *has at least one fixed point in* *M*.

In order to state another consequence of Theorem 2.1, the following abstract lemma is very useful.

**Lemma 2.1** *Assume that the conditions* (i), (ii) *and* (iv) *of Theorem* 2.1 *hold*. *If*, *moreover*, *S* *is sequentially weakly continuous and* $T(M)$ *is relatively weakly compact*, *then the set* $\mathcal{F}:={\mathcal{F}}^{(1,{x}_{0})}(T,S,M):=\{x\in M:x=Sx+Ty\phantom{\rule{0.25em}{0ex}}\mathit{\text{for some}}\phantom{\rule{0.25em}{0ex}}y\in M\}$ *is relatively weakly compact*.

*Proof*From the definition of ℱ it follows that

Since $0\le k<1$, then $w(\mathcal{F})=0$ and therefore ℱ is relatively compact. □

On the basis of Lemma 2.1, the following Krasnosel’skii-type fixed point theorem follows from Theorem 2.1.

**Corollary 2.3** [[8], Theorem 2.1]

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two sequentially weakly continuous mappings satisfying*:

- (i)
$T(M)$

*is relatively weakly compact*, - (ii)
*S**is a strict contraction*, - (iii)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

Now we consider the case where *S* is nonexpansive.

**Theorem 2.2**

*Let*

*E*

*be a Banach space and*

*μ*

*be a measure of weak noncompactness on E*.

*Let*

*M*

*be a nonempty bounded closed convex subset of*

*E*,

*and let*$T,S:M\to X$

*be two sequentially weakly continuous mappings satisfying*:

- (i)
*there are an integer**n**and a vector*${x}_{0}\in E$*such that**T**is**S*-*power*-*convex condensing w*.*r*.*t*.*μ*, - (ii)
*S**is a nonexpansive mapping*, - (iii)
*if*$({x}_{n})$*is a sequence of**M**such that*$((I-S){x}_{n})$*is weakly convergent*,*then the sequence*$({x}_{n})$*has a weakly convergent subsequence*, - (iv)
$I-S$

*is injective and demiclosed*, - (v)
$Tx+Sy\in M$

*for all*$x,y\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

*Proof*Let $z\in T(M)$. The map which assigns to each $x\in M$ the value $Sx+z$ defines a nonexpansive mapping from

*M*into

*M*. In view of Theorem 1.3, there exists a sequence $({x}_{n})$ in

*M*such that

*τ*the map which assigns to each $y\in M$ the point $\tau (y)\in M$ such that $(I-S)\tau (y)=Ty$. Since $I-S$ is injective, then $\tau :M\to M$ is well defined. It is easily seen that

*n*. Using the properties of the measure of weak noncompactness, we get

*C*is weakly compact. The reasoning in Theorem 2.1 shows that $\tau :C\to C$ is sequentially weakly continuous. Applying the Arino-Gautier-Penot fixed point theorem, we infer that there exists $x\in C$ such that

□

An easy consequence of Theorem 2.2 is the following.

**Corollary 2.4**

*Let*

*M*

*be a nonempty bounded closed convex subset of a reflexive Banach space*

*E*,

*and let*

*μ*

*be a measure of weak noncompactness on*

*E*.

*Suppose that*$T,S:M\to E$

*are two continuous mappings satisfying*:

- (i)
*there are an integer**n**and a vector*${x}_{0}\in E$*such that**T**is**S*-*power*-*convex condensing w*.*r*.*t*.*μ*, - (ii)
*S**is nonexpansive*, - (iii)
$I-S$

*is injective and demiclosed*, - (iv)
$Tx+Sy\in M$

*for all*$x,y\in M$.

*Then* $S+T$ *has at least one fixed point in* *M*.

*Proof* Keeping in mind that every bounded subset in a reflexive Banach space is relatively weakly compact, the result follows from Theorem 2.2. □

**Corollary 2.5**

*Let*

*M*

*be a nonempty bounded closed convex subset of a uniformly convex Banach space*

*E*.

*Suppose that*$T,S:M\to E$

*are two continuous mappings satisfying*:

- (i)
*there are an integer**n**and a vector*${x}_{0}\in E$*such that**T**is**S*-*power*-*convex condensing w*.*r*.*t*.*μ*, - (ii)
*S**is nonexpansive and*$I-S$*is injective*, - (iii)
$Tx+Sy\in M$

*for all*$x,y\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

*Proof* Note that in a uniformly convex space we have *S* is nonexpansive implies $I-S$ is demiclosed (see [21]). Moreover, every uniformly convex Banach space is reflexive. The result follows from Corollary 2.4. □

- (H1)
If ${({x}_{n})}_{n\in \mathbb{N}}$ is a weakly convergent sequence in $D(T)$, then ${(T{x}_{n})}_{n\in \mathbb{N}}$ has a strongly convergent subsequence in

*E*. - (H2)
If ${({x}_{n})}_{n\in \mathbb{N}}$ is a weakly convergent sequence in $D(T)$, then ${(T{x}_{n})}_{n\in \mathbb{N}}$ has a weakly convergent subsequence in

*E*.

**Remark 2.1** Continuous mappings satisfying (H1) are called ws-compact mappings and continuous mappings satisfying (H2) are called ww-compact mappings. Note that ws-compact and ww-compact mappings are not necessarily weakly continuous [12].

Now we state the following result.

**Theorem 2.3**

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*,

*and let*

*μ*

*be a measure of weak noncompactness on*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two continuous mappings satisfying*:

- (i)
*T**verifies*(H1), - (ii)
*S**is a strict contraction verifying*(H2), - (iii)
*there are an integer*${n}_{0}$*and a vector*${x}_{0}\in E$*such that**T**is**S*-*power*-*convex condensing w*.*r*.*t*.*μ**about*${x}_{0}$*and*${n}_{0}$, - (iv)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

*Proof*Let $y\in M$. The map which assigns to each $x\in E$ the value $Sx+Ty$ defines a strict contraction mapping from

*E*into itself and so it has a unique fixed point in

*E*by the contraction mapping principle [19]. Let us denote by $\tau :M\to E$ the map which assigns to each $y\in M$ the unique point in

*M*such that $\tau (y)=S\tau (y)+Ty$. From assumption (iv) we infer that $\tau (M)\subset M$. Notice

*n*we have

*C*is weakly compact. Put $D=\overline{\mathit{co}}(\tau (C))$ the closed convex hull of $\tau (C)$. We claim that

*D*is compact. Indeed, let $\{{x}_{n}\}$ be a sequence in

*C*. Since

*C*is weakly compact, then, up to a subsequence, we may assume that $\{{x}_{n}\}$ converges weakly to some $x\in C$. Bearing in mind that

*T*satisfies (H1), we infer that $\{T{x}_{n}\}$ has a norm convergent sequence, say $\{T{x}_{{n}_{k}}\}$, which converges to some $z\in E$. From the equality

*p*,

*q*we have

*E*and hence it is convergent. Thus, $\tau (C)$ is relatively compact and therefore $D=\overline{\mathit{co}}(\tau (C))$ is compact (see [22]). We now claim that $\tau :D\to D$ is continuous. Indeed, let $\{{x}_{n}\}$ be a sequence in

*D*such that ${x}_{n}\to x$ in

*D*. The continuity of

*T*guarantees that $T{x}_{n}\to Tx$. Hence,

□

The case $B=0$ in Theorem 2.3 corresponds to the following result, which was proved in [[12], Theorem 2.1].

**Corollary 2.6** *Let* *E* *be a Banach space and* *μ* *be a measure of weak noncompactness on* *E*. *Let* $M\subset E$ *be a nonempty closed convex and bounded subset*, ${x}_{0}\in E$, *and let* ${n}_{0}$ *be a positive integer*. *Suppose that* $T:M\to M$ *is power*-*convex condensing w*.*r*.*t*. *μ* *about* ${x}_{0}$ *and* ${n}_{0}$. *If* *T* *is ws*-*compact*, *then there exists* $x\in M$ *such that* $Tx=x$.

Another consequence of Theorem 2.3 is the following.

**Corollary 2.7**

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*,

*and let*

*μ*

*be a measure of weak noncompactness on*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two continuous mappings satisfying*:

- (i)
*T**verifies*(H1), - (ii)
*S**is a strict contraction verifying*(H2), - (iii)
*there are an integer*${n}_{0}$*and a vector*${x}_{0}$*such that*${\mathcal{F}}^{({n}_{0},{x}_{0})}(T,S,N)$*is relatively weakly compact*, - (iv)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

On the basis of Lemma 2.1 the following Krasnosel’skii-type fixed point theorem follows from Corollary 2.7.

**Corollary 2.8** [[15], Theorem 2.1]

*Let*

*M*

*be a nonempty bounded closed convex subset of a Banach space*

*E*.

*Suppose that*$T:M\to E$

*and*$S:E\to E$

*are two continuous mappings satisfying*:

- (i)
$T(M)$

*is relatively weakly compact and**T**satisfies*(H1), - (ii)
*S**is a strict contraction and**S**verifies*(H2), - (iii)
*if*$x=Sx+Ty$,*for some*$y\in M$,*then*$x\in M$.

*Then* $T+S$ *has at least one fixed point in* *M*.

## 3 Application

here $g:[0,T]\times X\to X$, $f:X\to X$ and ${x}_{0}\in X$ with *X* being a real Banach space. The integral in (3.1) is understood to be the Pettis integral and solutions to (3.1) will be sought in $E:=C([0,T],X)$.

- (i)
For each $t\in [0,T]$, ${g}_{t}=g(t,\cdot )$ is sequentially weakly continuous (

*i.e.*, for each $t\in [0,T]$, for each weakly convergent sequence $({x}_{n})$, the sequence ${g}_{t}({x}_{n})$ is weakly convergent). - (ii)
For each continuous $x:[0,T]\to E$, $g(\cdot ,x(\cdot ))$ is Pettis integrable on $[0,T]$.

- (iii)There exist $\alpha \in {L}^{1}[0,T]$ and $\theta :[0,+\mathrm{\infty})\to (0,+\mathrm{\infty})$ a nondecreasing continuous function such that $\parallel g(s,u)\parallel \le \alpha (s)\theta (\parallel u\parallel )$ for a.e. $s\in [0,T]$, and all $u\in X$, with${\int}_{0}^{T}\alpha (s)\phantom{\rule{0.2em}{0ex}}ds<{\int}_{0}^{\mathrm{\infty}}\frac{dx}{\theta (x)}.$
- (iv)There is a constant $\lambda \ge 0$ such that for any bounded subset
*S*of*X*and for any $t\in [0,T]$, we have$w\left(g([0,t]\times S)\right)\le \lambda w(S).$ - (v)
$f:X\to X$ is sequentially weakly continuous.

- (vi)
There exists $k\in [0,1)$ such that $\parallel f(u)-f(v)\parallel \le k\parallel u-v\parallel $ for all $u,v\in X$.

**Theorem 3.1** *Let* *X* *be a Banach space and suppose* (i)-(vi) *hold*. *Then* (3.1) *has a solution in* $E=C([0,T],X)$.

*Proof*Let

*M*is a closed, convex, bounded, equicontinuous subset of $C([0,T],X)$ with $0\in C$. To allow the abstract formulation of equation (3.1), we define the following operators $S,T:C([0,T],X)\to C([0,T],X)$ by

Our strategy is to apply Theorem 2.1 to show the existence of a fixed point for the sum $S+T$ in *M* which in turn is a continuous solution for equation (3.1). The proof will be divided into several steps.

Step 1: We show that $(x=Sx+Ty,y\in M)$ implies $x\in M$.

Consequently, $x\in M$.

*T*is

*S*-power-convex condensing w.r.t.

*w*about 0 and ${n}_{0}$, where

*w*is the De Blasi measure of weak noncompactness. To see this, notice, for each bounded set $\mathrm{\Omega}\subseteq M$ and for each $t\in [0,T]$, that

*M*is equicontinuous) that

*m*parts $0={t}_{0}<{t}_{1}<\cdots <{t}_{m}=t$ in such a way that $\mathrm{\Delta}{t}_{i}={t}_{i}-{t}_{i-1}=\frac{t}{m}$, $i=1,\dots ,m$. For each $x\in V$, we have

This gives the proof of the second step.

Step 3: *T* is sequentially weakly continuous. Let $\{{x}_{n}\}$ be a sequence in $C([0,T],X)$ such that ${x}_{n}\rightharpoonup x$ for some $x\in C([0,T],X)$. By Theorem 1.2 we have ${x}_{n}(t)\rightharpoonup x(t)$ in *X* for all $t\in [0,T]$. By assumption (i) we have $g(s,{x}_{n}(s))\rightharpoonup g(s,x(s))$ for all $s\in [0,T]$. The use of the Lebesgue dominated convergence theorem for Pettis integral [[23], Corollary 4] gives $(T{x}_{n})(t)\rightharpoonup (Tx)(t)$ for all $t\in [0,T]$. Using again Theorem 1.2, we obtain $T{x}_{n}\rightharpoonup Tx$. Thus, *T* is sequentially weakly continuous.

Applying Theorem 2.1, we get a fixed point for $S+T$ and hence a continuous solution to (3.1). □

## 4 Final remarks

## Declarations

### Acknowledgements

The authors would like to express their thanks to the editor and referees for their helpful comments and suggestions. This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under Grant No. (130-053-D1433). The authors, therefore, acknowledge with thanks DSR technical and financial support.

## Authors’ Affiliations

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