\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 39, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/39\hfil Nonlinear Volterra integral equations]
{Nonlinear Volterra integral equations in Henstock
integrability setting}

\author[B.-R. Satco\hfil EJDE-2008/39\hfilneg]
{Bianca-Renata Satco}

\address{Bianca-Renata Satco \newline
Faculty of Electrical Engineering and Computer Science,
``Stefan cel Mare'' University of Suceava \\
Universitatii 13 - Suceava - Romania}
\email{bisatco@eed.usv.ro}

\thanks{Submitted January 29, 2008. Published March 14, 2008.}
\thanks{Supported by grant 5954/18.09.2006 from MEdC - ANCS}
\subjclass[2000]{45D05, 26A39, 47H10}
\keywords{Nonlinear Volterra equation; Henstock integral;\hfill\break\indent
  Darbo's fixed point theorem}

\begin{abstract}
 We prove the existence of global solutions for nonlinear
 Volterra-type integral equations involving the Henstock integral
 in Banach spaces. The functions governing the equation are supposed
 to be continuous only with respect to some variables and to
 satisfy some  integrability or boundedness conditions.
 Our result improves  the similar one given in \cite{liuguowu}
 (where uniform continuity was required), as well as those referred
 therein. Then a related existence result is deduced in the
 set-valued case.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In the previous two decades, many authors showed  interest
in studying differential and integral problems under hypothesis of
integrability in a weaker sense than the classical ones;
by classical meaning Lebesgue integral on the real line, respectively
Bochner and Pettis integral in the vector case.
To be more precise, integrals of quite oscillating functions
were considered.
Thus, on the real line, we recall the results obtained using the
Henstock-Kurzweil integral (e.g.
\cite{chew flordeliza,federson bianconi,federson taboas,schwabik})
 and in the general case of Banach spaces, the
solutions given for similar problems under Henstock-Lebesgue
integrability assumptions (see
\cite{sikorska,sikorska1,sikorska3}),
in Henstock setting (e.g. \cite{satco1}) or
imposing some Henstock-Kurzweil-Pettis integrability conditions
(see \cite{cichon,satco3,sikorska2,sikorska3}).

 In the present paper, applying a Darbo-type fixed point
theorem established in \cite{liuguowu}, we obtain the existence of
global continuous solutions for the nonlinear Volterra integral
equation
$$
u(t)=(H)\int_0^t G(t,s)f\Big(s,u(s),\int_0^s
k(s,\tau)u(\tau)d\tau,\int_0^1 h(s,\tau)u(\tau)d\tau\Big)ds
$$
as well as for a related integral inclusion, considering the
Henstock integral (introduced in \cite{cao}). The setting is that
of a separable Banach space and the assumptions made on the
operators are much weaker than those previously imposed for
similar results (see \cite{liuguowu} and the papers cited there).
Mainly, we require some partial continuity to $f$ and $G$, along
with some integrability and boundedness conditions.

 Let us recall that
Volterra-type integral equations appearing in various nonlinear
problems in science were studied by many authors (see e.g.
\cite{guo,guolaksliu,liu,liuguowu,liuwuguo,oregan precup, oregan precup2})
via different fixed point theorems, such as Darbo-type theorems
(in \cite{guo,liuguowu}) or M\"{o}nch-type fixed point
theorems (in \cite{liu,oregan precup, oregan precup2}).


\section{Notation and preliminary facts}

Through this paper, $X$ is a real separable Banach space with norm
$\|\cdot\|$ and $R$-closed ball $T_R$ and $X^{\ast}$ is its
topological dual with unit ball $B^{\ast}$. $\mathcal{P}_0(X)$
stands for the family of nonempty subsets of $X$. $D$ denotes the
Pompeiu-Hausdorff distance and, for any $A\in \mathcal{P}_0(X)$,
$|A|=D(A,\{0\})$.

The space $C([0,1],X)$ of continuous functions
is endowed with the usual (Banach space) norm
$\|f\|_C={\sup_{t\in [0,1]}}\|f(t)\|$. Also,
$L^{\infty}([0,1],\mathbb{R})$ is the space of essentially bounded
real functions with essential supremum $\|\cdot\|_{L^{\infty}}$
and $BV([0,1],\mathbb{R})$ stands for the space of real bounded
variation function with its
classical norm $\|\cdot\|_{BV}$.

Let us now introduce the Henstock-type integrals in Banach spaces,
concepts that extend the real Henstock-Kurzweil integrability (for
which the reader is referred
 to \cite{gordon}). A gauge $\delta$ on $[0,1]$ is a
positive function. A partition of $[0,1]$ is a finite family
${(I_i,t_i)}_{i=1}^n$ of non-overlapping intervals covering
$[0,1]$ with the tags $t_i \in I_i$; a partition is said to be
$\delta$-fine if for each $i=1,\dots ,n$,
$I_i \subset ]t_i-\delta(t_i),t_i+\delta(t_i)[$.

\begin{definition}\label{henst} \rm
(see \cite{cao})
\begin{itemize}
\item[(i)] A function $f:[0,1]\to X$ is called
Henstock-integrable on $[0,1]$ if there exists an element
${\rm (H)}\int_0^1 f(s)ds\in X$ such that, for every
$\varepsilon >0$, there is a gauge $\delta_{\varepsilon}$ with
$$
\big\| \sum_{1=1}^n f(t_i)\mu(I_i)- {\rm (H)}\int_0^1 f(s)ds \big\|
<\varepsilon
$$
for every $\delta_{\varepsilon}$-fine partition;

\item[(ii)] $f$ is called Henstock-Lebesgue-integrable
(shortly, HL-integrable) on $[0,1]$ if there exists
$\widetilde{f}:[0,1]\to X$ such that, for every $\varepsilon > 0$,
there is a gauge $\delta_{\varepsilon} > 0$ satisfying, for each
$\delta_{\varepsilon}$-fine partition
${((x_{i-1},x_i),t_i)}_{i=1}^n$,
$$
\sum_{i=1}^n \big\|f(t_i)(x_i-x_{i-1})-\big[
\widetilde{f}(x_i)-\tilde{f}(x_{i-1}) \big] \big\| <
\varepsilon.
$$
 In this case, we denote $\widetilde{f}(t)={\rm
(HL)}\int_0^t f(s)ds$ and call it the HL-integral of $f$ on
$[0,t]$.
\end{itemize}
\end{definition}

\begin{remark} \label{rmk2} \rm
The Henstock and the Henstock-Lebesgue integrability are preserved
on sub-intervals of $[0,1]$, but not on any measurable subset. As about the
relationship between this integral and the classical ones, it is
well known that:
\begin{itemize}
\item[(i)] Every Bochner integrable function is HL-integrable, and the
converse is not valid;

\item[(ii)] the HL-integrability implies the Henstock integrability;

\item[(iii)] following a result given in \cite{fremlin mendoza},
any Pettis integrable function is Henstock integrable, but the
implication in the other sense is not true;

\item[(iv)] there exist Henstock-Lebesgue-integrable functions that are
not Pettis integrable (see the real case) and vice-versa (as
Example $42$ in {\rm \cite{gordon89}} shows).
\end{itemize}
In finite dimensional spaces, the two notions (of Henstock and
Henstock-Lebesgue integral) are equivalent. In the particular
real case, the preceding definition gives the Henstock-Kurzweil
(shortly, HK-) integral.
\end{remark}

The space of Henstock integrable $X$-valued functions is
denoted by  $\mathcal{H}([0,1],X)$, and is
equipped with the Alexiewicz norm:
$$
\| f\|_A=\sup_{[a,b]\subset [0,1]}\big\| (\text{H})\int_a^b
f(s)ds \big\|.
$$
It was proved that:

\begin{lemma}[\cite{saks}] \label{fctnala}
$T$ is a linear continuous functional on
the space of real HK-integrable functions if and only if there
exists a real function $g$ of bounded variation such that, for
every HK-integrable function $f$,
$T(f)={\rm (HK)}\int_0^1 f(s)g(s)ds$.
\end{lemma}

In order to extend this property to the vector Henstock integral,
let us recall several notions and results.

\begin{definition} \label{def4} \rm
A family $\mathcal{F}$ of HK-integrable functions is said to be
uniformly HK-integrable if, for each $\varepsilon >0$ there exists
a gauge $\delta_{\varepsilon}$ such that for every
$\delta_{\varepsilon}$-fine partition and every $f\in
\mathcal{F}$, $\big| \sum_{i=1}^n f(t_i)\mu(I_i)-({\rm
HK})\int_0^1 f(t)dt \big|
 < \varepsilon$.
\end{definition}

\begin{remark} \label{rmk6} \rm
The uniform HK-integrability is sometimes called
HK-equi-integrabi- lity (e.g.  \cite[Definition 3.5.1]{schwabik
guoju}).
\end{remark}

The following auxiliary result can be found in \cite{satco3}.

\begin{lemma}\label{bvunifhk}
Let $(f_n)_n$ be an uniformly HK-integrable, pointwisely bounded
sequence of real functions and $g$ be a function of bounded
variation. Then the sequence $(gf_n)_n$ is uniformly
HK-integrable.
\end{lemma}

A consequence of Proposition 1 in \cite{di piazza musial} is the
characterization result given in the sequel.

\begin{proposition} \label{propcaract}
If $f:[0,1]\to X$ is
scalarly HK-integrable (i.e. for any $x^{\ast}\in X^{\ast} $,
$\langle x^{\ast},f(\cdot) \rangle$
is HK-integrable), then the following conditions are equivalent:
\begin{itemize}
\item[(i)] $f$ is Henstock-integrable;

\item[(ii)] the collection $\{ \langle x^{\ast},f(\cdot)
\rangle;x^{\ast}\in B^{\ast} \}$ is uniformly HK-integrable;

\item[(iii)] each countable subset of $\{ \langle x^{\ast},f(\cdot)
\rangle;x^{\ast}\in B^{\ast} \}$ is uniformly HK-integrable.
\end{itemize}
\end{proposition}

 From here, one deduces the following result.

\begin{proposition} \label{prop8}
If $f:[0,1]\to X$ is Henstock-integrable and
$g:[0,1]\to \mathbb{R}$ is of bounded variation, then $fg$
is Henstock-integrable.
\end{proposition}

In the set-valued case, an Aumann-type integral will be
considered, via Henstock-integrable selections.

\begin{definition} \label{ef9} \rm
A set-valued function $F:[0,1]\to \mathcal{P}_0(X)$ is
Aumann-Henstock integrable if the collection of its
Henstock-integrable selections $S^H_F$ is non-empty. In this case,
the Aumann-Henstock integral of $F$ is defined by
$$
(AH)\int_0^1 F(s)ds=\big\{ (H)\int_0^1 f(s)ds, f\in S^H_F
\big\}
$$
\end{definition}

The Hausdorff measure of non-compactness $\alpha$ will play an
essential role in establishing the main results. For this, the
reader is referred to \cite{heinz}.

\begin{theorem}[\cite{ambrosetti}] \label{beta}
Let $\mathcal{K}\subset C([0,1],X)$ be bounded and equi-continuous.
Then $\alpha(\mathcal{K})={\sup_{t\in [0,1]}}
\alpha(\mathcal{K}(t))$.
\end{theorem}

The following result was proved in \cite{sikorska} for the
Henstock-Lebesgue integral, but the proof works for the Henstock
integral to. It is a generalization of some similar inequality
available for Bochner integrable functions, which can be found in
\cite{heinz}.

\begin{theorem}\label{hl}
If $M\subset \mathcal{H}([0,1],X)$ is a countable family
satisfying that, for some $h\in L^1([0,1],\mathbb{R})$,
$\alpha(M(t))\leq h(t)$ for a.e. $t\in [0,1]$, then
$\alpha(M(\cdot))\in L^1([0,1],\mathbb{R})$ and
$$
\alpha \Big((H)\int_0^t M(s)ds \Big) \leq \int_0^t
\alpha(M(s))ds, \quad \forall t\in [0,1].
$$
\end{theorem}

 The following generalization of Darbo's fixed point theorem was given
in \cite{liuguowu}.

\begin{lemma}\label{fix}
Let $F$ be a closed convex subset of a Banach space and the
operator $A:F\to F$ be continuous with $A(F)$ bounded.
Suppose that for the sequence defined for any bounded
$B \subset F$ by
$$
\tilde{A}^1(B)=A(B)\quad \text{and} \quad
\tilde{A}^n(B)=A\left(\overline{co}\left(\tilde{A}^{n-1}(B)\right)\right),
\quad \forall n\geq 2
$$
there exist a positive constant $0\leq k <1$ and a natural number
$n_0$ such that for every bounded $B \subset F$,
$\alpha(\tilde{A}^{n_0}(B))\leq k \alpha(B)$. Then $A$ has a fixed
point.
\end{lemma}



\section{Main results}

\begin{theorem}\label{principal}
Let $f:[0,1]\times X^3 \to X$ and $h,k,G:[0,1]^2 \to \mathbb{R}$
 satisfy the following conditions:
\begin{itemize}
\item[(i)] for each $t\in [0,1]$, $h(t,\cdot)$, $k(t,\cdot)$
are in $L^{\infty}([0,1],\mathbb{R})$, $G(t,\cdot)\in
BV([0,1],\mathbb{R})$, the application $t \mapsto G(t,\cdot)$ is
$\|\cdot\|_{BV}$-continuous and $t \mapsto h(t,\cdot)$,
$t \mapsto k(t,\cdot)$ are $L^{\infty}$-bounded;

\item[(ii)] for any $x,y,z\in C([0,1],X)$,
$f\left(\cdot,x(\cdot),y(\cdot),z(\cdot)\right)$ is
Henstock-integrable, and

\item[(ii)(1)] for each $R>0$ and $\varepsilon>0$, one can
find $\delta_{\varepsilon,R} >0$ such that
$$
\big\|(H)\int_{t_1}^{t_2}f(s,x,y,z)ds \big\|\leq \varepsilon,
\quad \forall \;|t_1-t_2|\leq \delta_{\varepsilon,R},\; \;
\forall x,y,z\in T_R
$$
 and
\begin{align*}
&\limsup_{R\to \infty}\frac{1}{R\delta_{1,R}}\\
&<\frac{1}{{\sup_{t\in[0,1]}}
\|G(t,\cdot)\|_{BV}\max\big\{ 1,\sup_{t\in[0,1]}
\|k(t,\cdot)\|_{L^{\infty}},\sup_{t\in[0,1]}
\|h(t,\cdot)\|_{L^{\infty}} \big\} };
\end{align*}

\item[(ii)(2)] the application
$(x,y,z)\mapsto f(\cdot,x,y,z)$ from $X^3$ towards
$\mathcal{H}([0,1],X)$ is $\|\cdot\|_A$-uniformly continuous;

\item[(iii)] there exist three positive integrable functions
such that for any bounded $D_i\subset X$ and any $t\in [0,1]$,
$$
\alpha\left(f(t,D_1,D_2,D_3)\right)\leq \sum_{i=1}^3
L_i(t)\alpha(D_i).
$$
\end{itemize}
Then the Volterra-type integral equation
$$
u(t)=(H)\int_0^t G(t,s)f\Big(s,u(s),\int_0^s
k(s,\tau)u(\tau)d\tau,\int_0^1 h(s,\tau)u(\tau)d\tau \Big)ds
$$
has a continuous solution on $[0,1]$.
\end{theorem}

\begin{proof} We follow the ideas of proof  in
\cite[Theorem 3.1]{liuguowu}. To simplify the calculus, denote by
$$
(Tu)(t)=\int_0^t k(t,s)u(s)ds, \quad
(Su)(t)=\int_0^1 h(t,s)u(s)ds.
$$
By hypothesis (i), let
$a={\sup_{t\in[0,1]}}\|k(t,\cdot)\|_{L^{\infty}}$,
$b={\sup_{t\in[0,1]}} \|G(t,\cdot)\|_{BV}$ and
$c={\sup_{t\in[0,1]}}\|h(t,\cdot)\|_{L^{\infty}}$.
One can find $0< r<{\frac{1}{b\max\{ 1,a,c \}}}$ and
$R_0>0$ such that for any $R\geq R_0 \max\{ 1,a,c \}$,
$$
\frac{1}{\delta_{1,R}}<rR.$$ Consider $A:C([0,1],X)\to C([0,1],X)$ defined
by
$$
Au(t)=(H)\int_0^t G(t,s)f(s,u(s),(Tu)(s),(Su)(s))ds.
$$
 We claim
that $A$ is a continuous operator applying the closed ball
$B_{R_0}$ of $C([0,1],X)$ into itself. Indeed, from (ii)(1), for
any $u \in C([0,1],X)$ with $\|u\|_C \leq R_0$,
\begin{align*}
\|Au\|_C &\leq {\sup_{t\in[0,1]}
\|G(t,\cdot)\|_{BV}\big\|(H)\int_0^t
f(s,u(s),(Tu)(s),(Su)(s))ds\big\|}\\
&\leq b \frac{1}{\delta_{1,R_0\max\{ 1,a,c\}}},
\end{align*}
since $\|Tu\|_C \leq aR_0$ and $\|Su\|_C \leq
cR_0$. It follows that
$$
\|Au\|_C <brR_0\max\{ 1,a,c\}<R_0.
$$
About the continuity, from hypothesis (ii)(2), one obtains that for every
$\varepsilon>0$ there is $\delta_{\varepsilon}>0$ such that
$$
\sup_{t\in[0,1]}\big\|(H)\int_0^t
f(s,x_1,y_1,z_1)-f(s,x_2,y_2,z_2)ds\big\|<\frac{\varepsilon}{b}
$$
for any $x_i,y_i$ satisfying $\max \{ \|x_1-x_2\|, \|y_1-y_2\|,
\|z_1-z_2\| \}<\max\{1,a,c\}\delta_{\varepsilon}$. Then
\begin{align*}
\|Au_1-Au_2\|_{C} &= \sup_{t\in[0,1]}\|Au_1(t)-Au_2(t)\|\\
&=\sup_{t\in[0,1]}\Big\| (H)\int_0^t
G(t,s)\Big(f(s,u_1(s),(Tu_1)(s),(Su_1)(s))\\
&\quad -f(s,u_2(s),(Tu_2)(s),(Su_2)(s))\Big)ds\Big\|\\
& \leq b\sup_{t\in[0,1]}\Big\| (H)\int_0^t
f(s,u_1(s),(Tu_1)(s),(Su_1)(s))\\
&\quad -f(s,u_2(s),(Tu_2)(s),(Su_2)(s)) ds\Big\|.
\end{align*}
If $\|u_1-u_2\|_C<\delta_{\varepsilon}$, then
$\|u_1(s)-u_2(s)\|<\delta_{\varepsilon}$,
$\|(Tu_1)(s)-(Tu_2)(s)\|<a\delta_{\varepsilon}$ and
$\|(Su_1)(s)-(Su_2)(s)\|<~c\delta_{\varepsilon}$, thus
$$
\|Au_1-Au_2\|_{C}<\varepsilon.
$$
Next, we prove that $F=\overline{co}A(B_{R_0})$ is
equi-continuous. Using  \cite[Lemma2.1]{liuguowu}, it is sufficient
to show that $A(B_{R_0})$ is equi-continuous.
For all $u\in B_{R_0}$ and all $0\leq t_1<t_2\leq 1$, we have
\begin{align*}
 \|Au(t_1)-Au(t_2)\|
& =\|(H)\int_0^{t_1}
G(t_1,s)f(s,u(s),(Tu)(s),(Su)(s))ds-\\
&\quad -(H)\int_0^{t_2}
G(t_2,s)f(s,u(s),(Tu)(s),(Su)(s))ds\|\\
& \leq \big\|(H)\int_0^{t_1}
(G(t_1,s)-G(t_2,s))f(s,u(s),(Tu)(s),(Su)(s))ds\big\|\\
& \quad +\big\|(H)\int_{t_1}^{t_2}
G(t_2,s)f(s,u(s),(Tu)(s),(Su)(s))ds\big\|\\
& \leq
\|G(t_1,\cdot)-G(t_2,\cdot)\|_{BV}\frac{1}{\delta_{1,R_0\max\{
1,a,c\}}}\\
& \quad +b{\sup_{t_1\leq t_1'<t_2'\leq
t_2}}\big\|(H)\int_{t_1'}^{t_2'}f(s,u(s),(Tu)(s),(Su)(s))ds\big\|
\end{align*}
and, by hypothesis (i) and (ii)(1), this can be made less than
some fixed $\varepsilon$
for $t_1, t_2$ with an appropriately small distance between them.
So, the equi-continuity follows.

Obviously, $A:F\to F$ is bounded and continuous.
Let us prove, by  mathematical induction, that for
every $B\subset F$ and any $n\in \mathbb{N}$,
$\tilde{A}^n(B)\subset A(B_{R_0})$, so $\tilde{A}^n(B)$ is bounded
and equi-continuous. For $n=1$, this is valid, since $A(B)\subset
A(F) \subset A(B_{R_0}) $. Suppose now that this is true for $n-1$
and prove it for $n$:
$$
\tilde{A}^n(B)=A(\overline{co}(\tilde{A}^{n-1}(B)))\subset
A(\overline{co}(A(B_{R_0})))\subset
A(\overline{co}(B_{R_0}))=A(B_{R_0}).
$$
Then, by Theorem
\ref{beta},
$$
\alpha\left(\tilde{A}^n(B)\right)=\sup_{t\in[0,1]}
\alpha\left(\tilde{A}^n(B)(t)\right), \quad \forall n\in \mathbb{N}.
$$
Similarly to the second part of the proof of Theorem 3.1 in
\cite{liuguowu}, one can show that there exist a constant $0\leq k
< 1 $ and a positive integer $n_0$ such that for any $B\subset
F$, $\alpha(\tilde{A}^{n_0}(B))\leq k \alpha(B)$.

Let $(v_n)_n$ be an arbitrary countable subset of
$\tilde{A}^1(B)=A(B)$. There exists a sequence $(u_n)_n \subset B$
such that $v_n=Au_n$. Hypothesis (ii)(1) allows us to use Theorem
\ref{hl} in order to obtain that
\begin{align*}
\alpha\left(\{v_n(t),n\in \mathbb{N}\}\right)
&=\alpha\big(\{Au_n(t),n\in \mathbb{N}\}\big)\\
&=\alpha\Big(\int_0^t
G(t,s)f(s,\{u_n(s),n\},\{(Tu_n)(s),n\},\{(Su_n)(s),n\})ds\Big)\\
& \leq b\int_0^t \alpha\left(
f(s,\{u_n(s),n\},\{(Tu_n)(s),n\},\{(Su_n)(s),n\}) \right)ds.
\end{align*}
By hypothesis (iii) it follows that
\begin{align*}
&\alpha(\{v_n(t),n\in \mathbb{N}\})\\
& \leq b\int_0^t L_1(s)\alpha(\{u_n(s) \}
)+L_2(s)\alpha(\{(Tu_n)(s)\})+L_3(s)\alpha(\{(Su_n)(s)\})ds.
\end{align*}
and, applying again Theorem \ref{hl},
\begin{align*}
\alpha(\{v_n(t),n\in \mathbb{N}\})
&\leq b\int_0^t
\left(L_1(s)+aL_2(s)+cL_3(s)\right)\alpha(\{u_n(s) \} )ds\\
& \leq  b\int_0^t \left(L_1(s)+aL_2(s)+cL_3(s))\right)ds
 \; \alpha(B).
\end{align*}
Since the Banach space is separable and the Hausdorff measure of
non-compactness is preserved when the set under discussion is
replaced by its adherence, this implies that
$$
\alpha\left(\tilde{A}^1(B)(t) \right)\leq b\int_0^t
\left(L_1(s)+aL_2(s)+cL_3(s))\right)ds \; \alpha(B).
$$
As $L_1(s)+aL_2(s)+cL_3(s)\in L^1([0,1],\mathbb{R})$ and
continuous functions are dense in $L^1([0,1],\mathbb{R})$ with
respect to the usual norm, for any $\varepsilon>0$ one can make an
evaluation of the form $\alpha(\tilde{A}^1(B)(t))\leq
(\varepsilon+Mt)\alpha(B)$. It can be shown, by mathematical
induction, that for every $m\in \mathbb{N}$,
$$
\alpha\left(\tilde{A}^{m}(B)(t) \right)\leq
\Big(\varepsilon^m+C_m^1\varepsilon^{m-1}Mt+\dots +\frac{(Mt)^m}{m!}
\Big)\alpha(B), \quad \forall t\in [0,1].
$$
Suppose the inequality is valid for $m$ and prove it for $m+1$. For any
countable subset $(v_n)_n$ of
$\tilde{A}^{m+1}(B)=A\left(\overline{co}\left(\tilde{A}^{m}(B)\right)
\right)$, there exist $(u_n)_n \subset
\overline{co}\left(\tilde{A}^{m}(B)\right)$ such that $v_n=Au_n$.
Then, as before,
$$
\alpha(\{v_n(t),n\in \mathbb{N}\}) \leq b\int_0^t
\left(L_1(s)+aL_2(s)+cL_3(s)\right)ds \;
\alpha\left(\tilde{A}^{m}(B)\right),
$$
whence
$$
\alpha\left(\tilde{A}^{m+1}(B)(t)\right) \leq b\int_0^t
\left(L_1(s)+aL_2(s)+cL_3(s)\right)ds \;
\alpha\left(\tilde{A}^{m}(B)\right)
$$
and so the assertion follows. The rest of the calculus goes as in
\cite{liuguowu}: for some integer $n_0$ the evaluation term
$\varepsilon^{n_0}+C_{n_0}^1\varepsilon^{{n_0}-1}Mt+\dots
+\frac{(Mt)^{n_0}}{{n_0}!}$
can be made less than $1$ and so, by Lemma \ref{fix}, $A$ has a
fixed point, which is a global solution to our equation.
\end{proof}

An existence result can be deduced in the set-valued case.

\begin{theorem}\label{principal-mult}
Let $F:[0,1]\times X^3 \to \mathcal{P}_{0}(X)$ and
$h,k,G:[0,1]^2 \to \mathbb{R}$ satisfy the following conditions:
\begin{itemize}
\item[(i)] for each $t\in [0,1]$, $h(t,\cdot)$, $k(t,\cdot)$
are in $L^{\infty}([0,1],\mathbb{R})$,
$G(t,\cdot)\in BV([0,1],\mathbb{R})$, the application
$t \mapsto G(t,\cdot)$ is
$\|\cdot\|_{BV}$-continuous and $t \mapsto h(t,\cdot)$,
$t \mapsto k(t,\cdot)$ are $L^{\infty}$-bounded;

\item[(ii)] for any $x,y,z\in C([0,1],X)$,
$F\left(\cdot,x(\cdot),y(\cdot),z(\cdot)\right)$ is
Aumann-Henstock integrable, and

\item[(ii)(1)] for each $R>0$ and $\varepsilon>0$, one can
find $\delta_{\varepsilon,R} >0$ such that
$$
\big|(AH)\int_{t_1}^{t_2}F(s,x,y,z)ds \big|
\leq \varepsilon, \quad \forall
\;|t_1-t_2|\leq \delta_{\varepsilon,R},\; \;  \forall x,y,z\in T_R
$$
 and
\begin{align*}
&\limsup_{R\to \infty}\frac{1}{R\delta_{1,R}}\\
&<\frac{1}{{\sup_{t\in[0,1]}}
\|G(t,\cdot)\|_{BV}\max\big\{ 1,\sup_{t\in[0,1]}
\|k(t,\cdot)\|_{L^{\infty}},\sup_{t\in[0,1]}
\|h(t,\cdot)\|_{L^{\infty}} \big\} };
\end{align*}

\item[(ii)(2)] the application $(x,y,z)\mapsto
S^H_{F(\cdot,x,y,z)}$ from $X^3$ towards
$\mathcal{P}_0(\mathcal{H}([0,1],X))$ possess $\|\cdot\|_A$-uniformly
 continuous selections;

\item[(iii)] there exist three positive integrable functions
such that for any bounded $D_i\subset X$ and any $t\in [0,1]$,
$$
\alpha\left(F(t,D_1,D_2,D_3)\right)\leq \sum_{i=1}^3
L_i(t)\alpha(D_i).
$$
\end{itemize}
Then the integral inclusion
$$
u(t)\in (AH)\int_0^t G(t,s)F\Big(s,u(s),\int_0^s
k(s,\tau)u(\tau)d\tau,\int_0^1 h(s,\tau)u(\tau)d\tau \Big)ds
$$
has global continuous solutions.
\end{theorem}

\begin{proof}
Every $\|\cdot \|_A$-uniformly continuous selection $f$ of
$(x,y,z)\mapsto S^H_{F(\cdot,x,y,z)}$ satisfies the hypothesis of
Theorem \ref{principal}, and so, the integral equation
$$
u(t)=(H)\int_0^t G(t,s)f\Big(s,u(s),\int_0^s
k(s,\tau)u(\tau)d\tau,\int_0^1 h(s,\tau)u(\tau)d\tau \Big)ds
$$
has a continuous solution on $[0,1]$. It follows that our
inclusion possess continuous solutions.
\end{proof}

\begin{remark}  \label{rmk15}\rm
 Theorems \ref{principal} and  \ref{principal-mult}
improve the related results given in
 \cite{guo,guolaksliu, liu, liuguowu,liuwuguo},
where the involved functions are supposed to be
uniformly continuous with respect to all arguments.
\end{remark}

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\end{document}