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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 31, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/31\hfil Solvability of a quadratic integral equation]
{Solvability of a quadratic integral equation of Fredholm type
in H\"older spaces}

\author[J. Caballero, M. A. Darwish, K. Sadarangani\hfil EJDE-2014/31\hfilneg]
{Josefa Caballero, Mohamed Abdalla Darwish,  Kishin Sadarangani}  % in alphabetical order

\address{Josefa Caballero \newline
Departamento de Matem\'aticas,
Universidad de Las Palmas de Gran Canaria,
Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain}
\email{fefi@dma.ulpgc.es}

\address{Mohamed Abdalla Darwish \newline
Department of Mathematics, Sciences Faculty for Girls,
 King Abdulaziz University, Jeddah, Saudi Arabia.\newline
Department of Mathematics, Faculty of Science,
Damanhour University, Damanhour, Egypt}
\email{dr.madarwish@gmail.com}

\address{Kishin Sadarangani \newline
Departamento de Matem\'aticas,
Universidad de Las Palmas de Gran Canaria,
Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain}
\email{ksadaran@dma.ulpgc.es}

\thanks{Submitted October 31, 2013. Published January 27, 2014.}
\subjclass[2000]{45G10, 45M99, 47H09}
\keywords{Fredholm; H\"older condition;
Schauder fixed point theorem}

\begin{abstract}
 In this article, we prove the existence of solutions of a quadratic integral
 equation of Fredholm type with a modified  argument, in the space
 of functions satisfying a H\"older condition.
 Our main tool is the classical Schauder fixed point theorem. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Differential equations with a modified arguments arise in a wide variety 
of scientific and technical applications, including the
modelling of problems from the natural and social sciences such as physics,
biological and economics sciences. A special class of these
differential equations have linear modifications of their arguments,
and have been studied by several authors, see \cite{Ba}--\cite{Mu2} and their
references. 

The aim of this article is to investigate the existence of solutions 
of the following integral equation of Fredholm type with a modified 
argument,
\begin{equation}\label{e1.1}
x(t)=p(t)+x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau,\quad t\in[0,1].
\end{equation}
Our solutions are placed in the space of functions satisfying the H\"older 
condition. A sufficient condition for the relative compactness
in these spaces and the classical Schauder fixed point theorem are 
the main tools in our study.

\section{Preliminaries}

Our starting point in this section is to introduce the space of functions 
satisfying the H\"older condition and some properties
in this space. These properties can be found in \cite{BaNa}.

Let $[a,b]$ be a closed interval in $\mathbb{R}$, by $C[a,b]$ we denote 
the space of continuous functions on $[a,b]$ equipped with the
supremum norm; i.e., $\|x\|_\infty=\sup\{|x(t)|:t\in[a,b]\}$ for $x\in C[a,b]$. 
For $0<\alpha\leq 1$ fixed, by $H_\alpha[a,b]$ we will denote
the space of the real functions $x$ defined on $[a,b]$ and satisfying 
the H\"older condition; that is, those functions $x$ for which there
exists a constant $H_x^\alpha$ such that
\begin{equation}\label{e2.1}
|x(t)-x(s)|\leq H_x^\alpha|t-s|^\alpha,
\end{equation}
for all $t,s\in[a,b]$. It is easily proved that $H_\alpha[a,b]$ 
is a linear subspace of $C[a,b]$.

In the sequel, for $x\in H_\alpha[a,b]$, by $H_x^\alpha$ we will denote 
the least possible constant for which inequality \eqref{e2.1} is
satisfied. More precisely, we put
\begin{equation}\label{e2.2}
H_x^\alpha=\sup\big\{\frac{|x(t)-x(s)|}{|t-s|^\alpha}:t\in[a,b],\;t\neq s\big\}.
\end{equation}
The spaces $H_\alpha[a,b]$ with $0<\alpha\leq 1$ can be equipped with the norm
$$
\|x\|_\alpha=|x(a)|+\sup\big\{\frac{|x(t)-x(s)|}{|t-s|^\alpha}:t\in[a,b],
\;t\neq s\big\},
$$
for $x\in H_\alpha[a,b]$. In \cite{BaNa}, the authors proved that 
$(H_\alpha[a,b],\|\cdot\|_\alpha)$ with $0<\alpha\leq 1$ is
a Banach space. The following lemmas appear in \cite{BaNa}.

\begin{lemma}\label{l2.1}
For $x\in H_\alpha[a,b]$ with $0<\alpha\leq 1$, the following inequality 
is satisfied
\begin{equation}\label{e2.3}
\|x\|_\infty\leq \max\big(1,(b-a)^\alpha\big)\|x\|_\alpha.
\end{equation}
\end{lemma}

\begin{lemma}\label{l2.2}
For $0<\alpha<\gamma\leq 1$, we have
\begin{equation}\label{e2.4}
H_\gamma[a,b]\subset H_\alpha[a,b]\subset C[a,b].
\end{equation}
Moreover, for $x\in H_\gamma[a,b]$ the following inequality holds
\begin{equation}\label{e2.5}
\|x\|_\alpha\leq \max\big(1,(b-a)^{\gamma-\alpha}\big)\|x\|_\gamma.
\end{equation}
\end{lemma}

Now, we present the following sufficient condition for relative compactness 
in the spaces $H_\alpha[a,b]$ with
$0<\alpha\leq 1$ which appears in Example $6$ of \cite{BaNa} and it is
 an important result for our study.

\begin{theorem}\label{t2.3}
Suppose that $0<\alpha<\beta\leq 1$ and that $A$ is a bounded subset 
in $H_\beta[a,b]$ (this means that $\|x\|_\beta\leq M$
for certain constant $M>0$, for any $x\in A$) then $A$ is a relatively 
compact subset of $H_\alpha[a,b]$.
\end{theorem}

\section{Main results}

In this section, we will study the solvability of \eqref{e1.1} in 
the H\"older spaces.
We will use the following assumptions:
\begin{enumerate}
\item[(i)] $p\in H_\beta[0,1]$, $0<\beta\leq 1$.

\item[(ii)] $k:[0,1]\times[0,1]\to\mathbb{R}$ is a continuous 
function such that it satisfies the H\"older
condition with exponent $\beta$ with respect to the first variable,
 that is, there exists a constant $K_\beta$ such that
$$
|k(t,\tau)-k(s,\tau)|\leq K_\beta\;|t-s|^\beta,
$$
for any $t,s,\tau\in [0,1]$.

\item[(iii)] $r:[0,1]\to[0,1]$ is a measurable function.

\item[(iv)] The following inequality is satisfied
$$
\|p\|_\beta(2K+K_\beta)<\frac{1}{4},
$$
where the constant $K$ is defined by
$$
K=\sup\big\{\int_0^1 |k(t,\tau)|\,d\tau:t\in[0,1]\big\},
$$
which exists by (ii).
\end{enumerate}

\begin{theorem}\label{t3.1}
Under assumptions {\rm (i)--(iv)}, Equation \eqref{e1.1} has at least one 
solution belonging to the space $H_\alpha[0,1]$,
where $\alpha$ is arbitrarily fixed number satisfying $0<\alpha<\beta$.
\end{theorem}

\begin{proof} 
Consider the operator $\mathcal{T}$ defined on $H_\beta[0,1]$ by
$$
(\mathcal{T}x)(t)=p(t)+x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau,\quad
t\in[0,1].
$$
In the sequel, we will prove that $\mathcal{T}$ transforms the space 
$H_\beta[0,1]$ into itself. In fact, we take
$x\in H_\beta[0,1]$ and $t,s\in[0,1]$ with $t\neq s$. 
Then, by assumptions (i) and (ii), we obtain
\begin{align*}
&\frac{|(\mathcal{T}x)(t)-(\mathcal{T}x)(s)|}{|t-s|^\beta}\\
&=\frac{\big|p(t)+x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau-p(s)
 -x(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau\big|}{|t-s|^\beta}
\\
&\leq \frac{|p(t)-p(s)|}{|t-s|^\beta}
 +\frac{\big|x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
-x(s)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau\big|}{|t-s|^\beta}\\
&\quad +\frac{\big|x(s)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
-x(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau\big|}{|t-s|^\beta}\\
&\leq \frac{|p(t)-p(s)|}{|t-s|^\beta}+\frac{|x(t)-x(s)|}{|t-s|^\beta}
 \int_0^1 |k(t,\tau)|\;|x(r(\tau))|\,d\tau\\
&\quad +\frac{|x(s)|\int_0^1 |k(t,\tau)-k(s,\tau)|\,
 |x(r(\tau))|\,d\tau}{|t-s|^\beta}\\
&\leq \frac{|p(t)-p(s)|}{|t-s|^\beta}
 +\frac{|x(t)-x(s)|}{|t-s|^\beta}\|x\|_\infty\int_0^1 |k(t,\tau)|\,d\tau\\
&\quad +\frac{\|x\|_\infty\cdot\|x\|_\infty\int_0^1 |k(t,\tau)-k(s,\tau)|
 \,d\tau}{|t-s|^\beta}\\
&\leq \frac{|p(t)-p(s)|}{|t-s|^\beta}
 +K\|x\|_\infty\frac{|x(t)-x(s)|}{|t-s|^\beta}
 +\frac{\|x\|_\infty^2\int_0^1 K_\beta|t-s|^\beta\,d\tau}{|t-s|^\beta}\\
&\leq  H_p^\beta+K\|x\|_\infty H_x^\beta+K_\beta\|x\|_\infty^2.
\end{align*}
By Lemma \ref{l2.1}, since $\|x\|_\infty\leq\|x\|_\beta$ and, as 
$H_x^\beta\leq\|x\|_\beta$, we infer that
$$
\frac{|(\mathcal{T}x)(t)-(\mathcal{T}x)(s)|}{|t-s|^\beta}
\leq H_p^\beta+(K+K_\beta)\|x\|_\beta^2.
$$
Therefore,
\begin{equation}
\begin{aligned} \label{e3.1}
\|\mathcal{T}x\|_\beta
&=|(\mathcal{T}x)(0)|+\sup\big\{\frac{|(\mathcal{T}x)(t)-(\mathcal{T}x)(s)|}
{|t-s|^\beta}:t,s\in[0,1],t\neq s\big\}\\
&\leq |(\mathcal{T}x)(0)|+H_p^\beta+(K+K_\beta)\|x\|_\beta^2\\
&\leq |p(0)|+|x(0)|\int_0^1 |k(0,\tau)|\;|x(r(\tau))|\,d\tau
 +H_p^\beta+(K+K_\beta)\|x\|_\beta^2\\
&\leq \|p\|_\beta+\|x\|_\infty\cdot\|x\|_\infty\int_0^1 |k(0,\tau)|\,d\tau
 +(K+K_\beta)\|x\|_\beta^2\\
&\leq \|p\|_\beta+K\|x\|_\beta^2+(K+K_\beta)\|x\|_\beta^2\\
&=\|p\|_\beta+(2K+K_\beta)\|x\|_\beta^2<\infty.
\end{aligned}
\end{equation}
This proves that the operator $\mathcal{T}$ maps $H_\beta[0,1]$ into itself.

Taking into account that the inequality
$$
\|p\|_\beta+(2K+K_\beta)r^2<r
$$
is satisfied for values between the numbers
$$
r_1=\frac{1-\sqrt{1-4\|p\|_\beta(2K+K_\beta)}}{2(2K+K_\beta)}
$$
and
$$
r_2=\frac{1-\sqrt{1+4\|p\|_\beta(2K+K_\beta)}}{2(2K+K_\beta)}
$$
which are positive by assumption (iv), consequently, from \eqref{e3.1} 
it follows that $\mathcal{T}$ transforms the ball
$B_{r_0}^\beta=\{x\in H_\beta[0,1]:\|x\|_\beta\leq r_0\}$ into itself, 
for any $r_0\in[r_1,r_2]$; i.e.,
$\mathcal{T}:B_{r_0}^\beta\to B_{r_0}^\beta$, where $r_1\leq r_0\leq r_2$.

By Theorem \ref{t2.3}, we have that the set
$B_{r_0}^\beta$ is relatively compact in $H_\alpha[0,1]$ for any 
$0<\alpha<\beta\leq 1$. Moreover, we can prove that $B_{r_0}^\beta$
is a compact subset in $H_\alpha[0,1]$ for any $0<\alpha<\beta\leq 1$ 
(see Appendix).

Next, we will prove that the operator $\mathcal{T}$ is continuous on 
$B_{r_0}^\beta$, where in $B_{r_0}^\beta$ we consider the induced
norm by $\|\cdot\|_\alpha$, where $0<\alpha<\beta\leq 1$.
To do this, we fix $x\in B_{r_0}^\beta$ and $\varepsilon>0$. 
Suppose that $y\in B_{r_0}^\beta$ and $\|x-y\|_\alpha\leq\delta$, 
where $\delta$ is a positive number such that 
$\delta<\frac{\varepsilon}{2(2K+3K_\beta)r_0}$.

Then, for any $t,s\in[0,1]$ with $t\neq s$, we have
\begin{align*}
&\frac{|[(\mathcal{T}x)(t)-(\mathcal{T}y)(t)]-[(\mathcal{T}x)(s)-
(\mathcal{T}y)(s)]|}{|t-s|^\alpha}\\
&=\Big|\frac{\big[x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
 -y(t)\int_0^1 k(t,\tau)\;y(r(\tau))\,d\tau\big]}{|t-s|^\alpha}\\
&\quad -\frac{\big[x(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau
 -y(s)\int_0^1 k(s,\tau)\;y(r(\tau))\,d\tau\big]}{|t-s|^\alpha} \Big|\\
&\leq\Big| \frac{\big[x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
 -y(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau\big]}{|t-s|^\alpha}\\
&\quad +\frac{\big[y(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
 -y(t)\int_0^1 k(t,\tau)\;y(r(\tau))\,d\tau\big]}{|t-s|^\alpha}\\
&\quad -\frac{\big[x(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau
 -y(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau\big]}{|t-s|^\alpha}\\
&\quad -\frac{\big[y(s)\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau
 -y(s)\int_0^1k(s,\tau)\;y(r(\tau))\,d\tau\big]}{|t-s|^\alpha}\Big|\\
&=\frac{1}{|t-s|^\alpha}\Big|(x(t)-y(t))\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau
 +y(t)\int_0^1 k(t,\tau)(x(r(\tau)-y(r(\tau)))\,d\tau \\
&\quad -(x(s)-y(s))\int_0^1 k(s,\tau)\;x(r(\tau))\,d\tau
 -y(s)\int_0^1 k(s,\tau)\;(x(r(\tau))-y(r(\tau)))\,d\tau\Big|\\
&\leq\frac{1}{|t-s|^\alpha}\Big\{
|(x(t)-y(t))-(x(s)-y(s))|\cdot\Big|\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau\Big|\\
&\quad +|x(s)-y(s)|\cdot\Big|\int_0^1(k(t,\tau)-k(s,\tau))\;x(r(\tau))
 \,d\tau\dot|\\
&\quad +\Big|y(t)\int_0^1 k(t,\tau)(x(r(\tau)-y(r(\tau)))\,d\tau
 -y(s)\int_0^1 k(s,\tau)(x(r(\tau)-y(r(\tau)))\,d\tau\Big|\Big\}
\\ 
&\leq\frac{|(x(t)-y(t))-(x(s)-y(s))|}{|t-s|^\alpha}\|x\|_\infty
 \int_0^1 |k(t,\tau)|\,d\tau\\
&\quad  +\big[|(x(s)-y(s))-(x(0)-y(0))|+|x(0)-y(0)|\big]\|x\|_\infty\\
&\quad\times \int_0^1\frac{|k(t,\tau)-k(s,\tau)|}{|t-s|^\alpha}d\tau
 +\frac{1}{|t-s|^\alpha}\Big|y(t)\\
&\quad\times \int_0^1 k(t,\tau)(x(r(\tau)
 -y(r(\tau)))\,d\tau-y(s)\int_0^1 k(t,\tau)(x(r(\tau)-y(r(\tau)))\,d\tau\Big|\\
&\quad +\frac{1}{|t-s|^\alpha}\Big|
y(s)\int_0^1 k(t,\tau)(x(r(\tau)-y(r(\tau)))\,d\tau\\
&\quad -y(s)\int_0^1 k(s,\tau)
 (x(r(\tau)-y(r(\tau)))\,d\tau\Big|
\\
&\leq K \|x-y\|_\alpha\|x\|_\infty+\sup_{p,q\in[0,1]}|(x(p)-y(p))-(x(q)-y(q))|\\
&\quad\times \|x\|_\infty\int_0^1\frac{K_\beta|t-s|^\beta}{|t-s|^\alpha}d\tau
 +|x(0)-y(0)|\|x\|_\infty
\int_0^1\frac{K_\beta|t-s|^\beta}{|t-s|^\alpha}d\tau\\
&\quad +\frac{|y(s)-x(s)|}{|t-s|^\alpha}
 \int_0^1 |k(t,\tau)|\;|x(r(\tau)-y(r(\tau))|\,d\tau\\
&\quad +|y(s)|\int_0^1\frac{|k(t,\tau)-k(s,\tau)|}{|t-s|^\alpha}
|x(r(\tau)-y(r(\tau))|\,d\tau
\\
&\leq K\|x\|_\infty\|x-y\|_\alpha
 +\|x\|_\infty K_\beta|t-s|^{\beta-\alpha}\\
&\quad\times \sup_{p,q\in[0,1],\;p\neq q}
\big\{\frac{|(x(p)-y(p))-(x(q)-y(q))|}{|p-q|^\alpha}|p-q|^\alpha\big\}\\
&\quad +K_\beta\|x\|_\beta|t-s|^{\beta-\alpha}|x(0)-y(0)|\\
&\quad +K H_y^\alpha\|x-y\|_\infty+\|y\|_\infty\;\|x-y\|_\infty
\int_0^1\frac{K_\beta\;|t-s|^\beta}{|t-s|^\alpha}d\tau\\
&\leq K\|x\|_\beta\|x-y\|_\alpha+2K_\beta\|x\|_\beta\|x-y\|_\alpha+
K\|y\|_\alpha\|x-y\|_\alpha+K_\beta\|y\|_\alpha\|x-y\|_\alpha\\
&\leq\big(
K\|x\|_\beta+2K_\beta\|x\|_\beta+K\|y\|_\alpha+K_\beta\|y\|_\alpha\big)
\|x-y\|_\alpha.
\end{align*}
Since $\|y\|_\alpha\leq\|y\|_\beta$ (see, Lemma \ref{l2.2}) and 
$x,y\in B_{r_0}^\beta$, from the above inequality we infer that
\begin{equation} \label{e3.2}
\begin{aligned}
\frac{|[(\mathcal{T}x)(t)-(\mathcal{T}y)(t)]-[(\mathcal{T}x)(s)-
(\mathcal{T}y)(s)]|}{|t-s|^\alpha}
&\leq(2Kr_0+3K_\beta r_0)\|x-y\|_\alpha\\
&\leq(2Kr_0+3K_\beta r_0)\delta<\frac{\varepsilon}{2}.
\end{aligned}
\end{equation}
On the other hand,
\begin{equation} \label{e3.3}
\begin{aligned}
 |(\mathcal{T}x)(0)-(\mathcal{T}y)(0)|
&= \Big|x(0)\int_0^1k(0,\tau)\;x(r(\tau))\,d\tau-y(0)\int_0^1k(0,\tau)
  y(r(\tau))\,d\tau\Big|\\
&\leq \Big|x(0)\int_0^1k(0,\tau)\;x(r(\tau))\,d\tau
 -x(0)\int_0^1k(0,\tau)\;y(r(\tau))\,d\tau\Big|\\
&\quad +\Big|x(0)\int_0^1k(0,\tau)\;y(r(\tau))\,d\tau-y(0)
\int_0^1k(0,\tau)\;y(r(\tau))\,d\tau\Big|\\
&\leq \Big| x(0)\int_0^1k(0,\tau)(x(r(\tau))-y(r(\tau)))\,d\tau\Big|\\
&\quad +\Big| (x(0)-y(0))\int_0^1k(0,\tau)\;y(r(\tau)))\,d\tau\Big|\\
&\leq K\|x\|_\infty\|x-y\|_\infty+K\|y\|_\infty\|x-y\|_\alpha\\
&\leq K\|x\|_\beta\|x-y\|_\alpha+K\|y\|_\beta\|x-y\|_\alpha\\
&\leq  2Kr_0\|x-y\|_\alpha<2Kr_0\delta<\frac{\varepsilon}{2}.
\end{aligned}
\end{equation}
From \eqref{e3.2} and \eqref{e3.3}, it follows that
\begin{align*}
&\|\mathcal{T}x-\mathcal{T}y\|\\
&=|(\mathcal{T}x)(0)-(\mathcal{T}y)(0)|\\
&\quad +\sup\big\{ \frac{|((\mathcal{T}x)(t)-(\mathcal{T}y)(t))-
((\mathcal{T}x)(s)-(\mathcal{T}y)(s))|}{|t-s|^\alpha}:t,s\in[0,1],
\;t\neq s\big\}\\
&< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
\end{align*}
This proves that the operator $\mathcal{T}$ is continuous at the
point $x\in B_{r_0}^\delta$ for the norm $\|\|_\alpha$.
Since $B_{r_0}^\delta$ is compact in $H_\alpha[0,1]$, applying the
classical Schauder fixed point theorem we obtain the desired result.
\end{proof}

\section{Example}
To present an example illustrating our result we need some previous results.

\begin{definition}\label{d3.2} \rm
A function $f:\mathbb{R}_+\to \mathbb{R}_+$ is said to be subadditive 
if $f(x+y)\leq f(x)+f(y)$ for any $x,y\in\mathbb{R}_+$.
\end{definition}

\begin{lemma}\label{l3.3}
Suppose that $f:\mathbb{R}_+\to \mathbb{R}_+$ is subadditive and 
$y\leq x$ then $f(x)-f(y)\leq f(x-y)$.
\end{lemma}

\begin{proof} Since
$f(x)=f(x-y+y)\leq f(x-y)+f(y)$
the result follows.
 \end{proof}

\begin{remark}\label{r3.4} \rm
From Lemma \ref{l3.3}, we infer that if $f:\mathbb{R}_+\to \mathbb{R}_+$ 
is subadditive then $|f(x)-f(y)|\leq f(|x-y|)$
for any $x,y\in\mathbb{R}_+$.
\end{remark}

\begin{lemma}\label{l3.5}
Let $f:\mathbb{R}_+\to \mathbb{R}_+$ be a concave function with $f(0)=0$. 
Then $f$ is subadditive.
\end{lemma}

\begin{proof} 
For $x,y\in\mathbb{R}_+$ and, since $f$ is concave and $f(0)=0$, we have
\begin{align*}
f(x)&=f\left(\frac{x}{x+y}(x+y)+\frac{y}{x+y}\cdot 0\right)\\
&\geq\frac{x}{x+y}f(x+y)+\frac{y}{x+y}f(0)\\
&= \frac{x}{x+y}f(x+y)
\end{align*}
and
\begin{align*}
f(y)&=f\left(\frac{x}{x+y}\cdot 0+\frac{y}{x+y}(x+y)\right)\\
&\geq \frac{x}{x+y}f(0)f(x+y)+\frac{y}{x+y}f(x+y)\\
&= \frac{y}{x+y}f(x+y).
\end{align*}
Adding these inequalities, we obtain
$$
f(x)+f(y)\geq \frac{x}{x+y}f(x+y)+\frac{y}{x+y}f(x+y)=f(x+y).
$$
This completes the proof. 
\end{proof}

\begin{remark}\label{r3.6} \rm
Let $f:\mathbb{R}_+\to \mathbb{R}_+$ be the function defined by
 $f(x)=\sqrt[p]{x}$, where $p>1$. Since this function
is concave (because $f''(x)\leq 0$ for $x>0$) and $f(0)=0$, 
Lemma \ref{l3.5} says us that $f$ is subadditive. By Remark
\ref{r3.4}, we have
$$
|f(x)-f(y)|=|\sqrt[p]{x}-\sqrt[p]{y}|\leq \sqrt[p]{|x-y|}
$$
for any $x,y\in\mathbb{R}_+$.
\end{remark}

\begin{example}\rm
 Let us consider the  quadratic integral equation
\begin{equation}\label{e3.4}
x(t)=\arctan \sqrt[5]{q\sin t+\hat{q}}+x(t)\int_0^1 \sqrt[4]{m t^2
+\tau}\;x\big(\frac{\tau}{\tau+1}\big)\,d\tau,\quad t\in[0,1],
\end{equation}
where, $q$, $\hat{q}$ and $m$ are nonnegative constants.
Notice that \eqref{e3.4} is a particular case of \eqref{e1.1}, where 
$p(t)=\arctan \sqrt[5]{q\sin t+\hat{q}}$, $k(t,\tau)=\sqrt[4]{m t^2+\tau}$ and
$r(\tau)=\frac{\tau}{\tau+1}$.

In what follows, we will prove that assumptions (i)-(iv) of Theorem \ref{t3.1} 
are satisfied. Since the inverse tangent function is concave 
(because its second derivative is nonpositive) and its value at zero is zero, 
taking into account Remarks \ref{r3.4} and \ref{r3.6}, we have
\begin{align*}
|p(t)-p(s)|
&= \Big|\arctan \sqrt[5]{q\sin t+\hat{q}}
 -\arctan \sqrt[5]{q\sin s+\hat{q}}\Big|\\
&\leq \arctan\Big(\big|\sqrt[5]{q\sin t+\hat{q}}-\sqrt[5]{q\sin s+\hat{q}}\big|
 \Big)\\
&\leq \big|\sqrt[5]{q\sin t+\hat{q}}-\sqrt[5]{q\sin s+\hat{q}}\big|\\
&\leq \sqrt[5]{q|\sin t-\sin s|}\\
&\leq \sqrt[5]{q}\;|t-s|^{1/5},
\end{align*}
where we have used that $\arctan x\leq x$ for $x\geq 0$ and 
$|\sin x-\sin y| \leq |x-y|$ for any $x,y\in\mathbb{R}$.
 This says that $p\in H_{\frac{1}{5}}[0,1]$ and, moreover, 
$H_p^{1/5}=\sqrt[5]{q}$. Therefore,
assumptions (i) of Theorem \ref{t3.1} is satisfied.

Note that
\begin{align*}
\|p\|_{\frac{1}{5}}
&=|p(0)|+\sup\big\{ \frac{|p(t)-p(s)|}{|t-s|^{1/5}}:
 t,s\in[0,1],\;t\neq s\big\}\\
&\leq \arctan\sqrt[5]{\hat{q}}+H_p^{1/5}\\
&=\arctan\sqrt[5]{\hat{q}}+\sqrt[5]{q}.
\end{align*}
Since for any $t,s,\tau\in[0,1]$, we have (see, Remark \ref{r3.6})
\begin{align*}
|k(t,\tau)-k(s,\tau)|
&=\big|\sqrt[4]{m t^2+\tau}-\sqrt[4]{m s^2+\tau}\big|\\
&\leq \sqrt[4]{|m t^2-m s^2|}\\
&=\sqrt[4]{m}\sqrt[4]{|t^2-s^2|}\\
&=\sqrt[4]{m}\sqrt[4]{t+s}\sqrt[4]{|t-s|}\\
&\leq \sqrt[4]{m}\sqrt[4]{2}\;|t-s|^{1/4}\\
&=\sqrt[4]{m}\sqrt[4]{2}\;|t-s|^{1/20}\;|t-s|^{1/5}\\
&\leq \sqrt[4]{2m}\;|t-s|^{1/5},
\end{align*}
assumption $(ii)$ of Theorem \ref{t3.1} is satisfied with 
$K_\beta=K_{\frac{1}{5}}=\sqrt[4]{2m}$.

It is clear that $r(\tau)=\frac{\tau}{\tau+1}$ satisfies assumption (iii).

In our case, the constant $K$ is given by
\begin{align*}
K&=\sup\big\{\int_0^1 |k(t,\tau)|\,d\tau:t\in[0,1]\big\}\\
&=\sup\big\{\int_0^1 \sqrt[4]{m t^2+\tau}\,d\tau:t\in[0,1]\big\}\\
&=\sup\big\{\frac{4}{5}\Big(\sqrt[4]{(m t^2+1)^5}-\sqrt[4]{m^5 t^{10}}\Big)\big\}\\
&= \frac{4}{5}\Big(\sqrt[4]{(m+1)^5}-\sqrt[4]{m^5}\Big).
\end{align*}
Therefore, the inequality appearing in assumption $(iv)$ takes the form
$$
\|p\|_{\frac{1}{5}}(2K+K_\beta)
=\Big(\arctan\sqrt[5]{\hat{q}}+\sqrt[5]{q}\Big)
\Big(\frac{8}{5}\big[\sqrt[4]{(m+1)^5}-\sqrt[4]{m^5}\big]
+\sqrt[4]{2m}\Big)<\frac{1}{4}.
$$
It is easily seen that the above inequality is satisfied when, 
for example, $\hat q=0$, $q=\frac{1}{2^{20}}$ and $m=1$. Therefore, 
using Theorem \ref{t3.1}, we infer that \eqref{e3.4} for $\hat q=0$,
 $q=\frac{1}{2^{20}}$ and $m=1$ has at least one solution in the 
space $H_\alpha[0,1]$ with $0<\alpha<1/5$.
\end{example}

Note that in \eqref{e3.4}, we can take as $r(\tau)$ a particular functions 
such as $r(\tau)=\{e^\tau\}$, where $\{\cdot\}$ denotes the fractional part.

\begin{remark}\label{r3.8} \rm
Note that any solution $x(t)$ of \eqref{e1.1}, i.e.,
$$
x(t)=p(t)+x(t)\int_0^1 k(t,\tau)\;x(r(\tau))\,d\tau,\quad t\in[0,1],
$$
satisfies that its zeroes are also zeroes of $p(t)$. 
From this, we infer that if $p(t)\neq 0$ for any $t\in[0,1]$ then 
$x(t)\neq 0$ for any $t\in[0,1]$. By Bolzano's theorem, this means 
that the solution $x(t)$ of Eq\eqref{e1.1} does not change of sign 
on $[0,1]$ when $p(t)\neq 0$ for any $t\in[0,1]$.
These questions seem to be interesting from a practical standpoint.
\end{remark}

\section{Appendix}

Suppose that $0<\alpha<\beta\leq 1$ and by $B_r^\beta$ we denote the ball 
centered at $\theta$ and radius $r$ in the space $H_\beta[a,b]$; i.e.,
$B_r^\beta=\{x\in H_\beta[a,b]:\|x\|_\beta\leq r\}$. 
Then $B_r^\beta$ is a compact subset in the space $H_\alpha[a,b]$.

In fact, by Theorem
\ref{t2.3}, since $B_r^\beta$ is a bounded subset in $H_\beta[a,b]$, 
$B_r^\beta$ is a relatively compact subset of $H_\alpha[a,b]$. In the sequel,
we will prove that $B_r^\beta$ is a closed subset of $H_\alpha[a,b]$. 
Suppose that
$(x_n)\subset B_r^\beta$ and $x_n\stackrel{\|\|_\alpha}{\longrightarrow}x$ 
with $x\in H_\alpha[a,b]$. We have to prove that $x\in B_r^\beta$.

Since $x_n\stackrel{\|\cdot\|_\alpha}{\longrightarrow}x$, for $\varepsilon>0$ 
given we can find $n_0\in \mathbb{N}$ such that 
$\|x_0-x\|_\alpha\leq \varepsilon$ for any $n\geq n_0$, or, equivalently,
\begin{equation}\label{e3.5}
|x_n(a)-x(a)|+\sup\big\{\frac{|(x_n(t)-x(t))-(x_n(s)-x(s))|}{|t-s|^\alpha}
:t,s\in[a,b],\;t\neq s\big\}<\varepsilon,
\end{equation}
for any $n\geq n_0$. Particularly, this implies that $x_n(a)\to x(a)$. 
Moreover, if in \eqref{e3.5} we put $s=a$ then we get
$$
\sup\big\{\frac{|(x_n(t)-x(t))-(x_n(a)-x(a))|}{|t-a|^\alpha}
:t,s\in[a,b],\;t\neq a\big\}<\varepsilon,\;{\rm for}\;{\rm any}\;n\geq n_0.$$
This says that
\begin{equation}\label{e3.6}
|(x_n(t)-x(t))-(x_n(a)-x(a))|<\varepsilon|t-a|^\alpha,
\quad\text{for any $n\geq n_0$ and for any $t\in[a,b]$}.
\end{equation}
Therefore, for any $n\geq n_0$ and any $t\in[a,b]$ by \eqref{e3.5} 
and \eqref{e3.6}, we have
\begin{align*}
|x_n(t)-x(t)|
&\leq |(x_n(t)-x(t))-(x_n(a)-x(a))|+|x_n(a)-x(a)|\\
&<\varepsilon(t-a)^\alpha+\varepsilon\\
&=\varepsilon(1+(b-a)^\alpha).
\end{align*}
From this, it follows that
\begin{equation}\label{e3.7}
\|x_n-x\|_\infty\to 0.
\end{equation}
Next, we will prove that $x\in B_r^\beta$.
In fact, as $(x_n)\subset B_r^\beta\subset H_\beta[a,b]$, we have that
$$
\frac{|x_n(t)-x_n(s)|}{|t-s|^\beta}\leq r
$$
for any $t,s\in[a,b]$ with $t\neq s$. Consequently,
$$
|x_n(t)-x_n(s)|\leq r|t-s|^\beta
$$
for any $t,s\in[a,b]$. Letting $n\to\infty$ and taking into
account \eqref{e3.7}, we obtain
$$
|x(t)-x(s)|\leq r|t-s|^\beta
$$
for any $t,s\in[a,b]$. Therefore,
$$
\frac{|x(t)-x(s)|}{|t-s|^\beta}\leq r
$$
for any $t,s\in[a,b]$ with $t\neq s$, and this means that $x\in B_r^\beta$.
This completes  the proof.

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\end{document}