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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 133, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/133\hfil 
 Fixed point results in metric-like spaces]
{Fixed point results for generalized $\alpha$-$\psi$-contractions in
metric-like spaces and applications}

\author[H. Aydi, E. Karapinar \hfil EJDE-2015/133\hfilneg]
{Hassen Aydi, Erdal Karapinar} 

\address{Hassen Aydi \newline
Dammam University,
Departement of Mathematics,
College of Education of Jubail,  
P.O. 12020, Industrial Jubail 31961, Saudi Arabia}
\email{hmaydi@uod.edu.sa}

\address{Erdal Karapinar \newline
Atilim University, 
Department of Mathematics, 06836, \.Incek, Ankara, Turkey.\newline
Nonlinear Analysis and Applied Mathematics Research Group (NAAM), 
King Abdulaziz University, 21589,  Jeddah, Saudi Arabia}
\email{erdalkarapinar@yahoo.com,  ekarapinar@atilim.edu.tr}

\thanks{Submitted December 2, 2014. Published May 15, 2015.}
\subjclass[2010]{47H10, 54H25}
\keywords{Metric-like; $\alpha$-admissible mappings; fixed point} 

\begin{abstract}
 In this article, we introduce the concept of generalized
 $\alpha$-$\psi$-con\-traction in the context of metric-like spaces and
 establish some related fixed point theorems.
 As consequences, we obtain some known fixed point results in the literature.
 Some examples and an application on two-point boundary value problems for
 second order differential equation are also considered.
\end{abstract}

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


\section{Introduction and preliminaries}

The notion of metric-like (dislocated) metric spaces was introduced by
Hitzler and Seda \cite{HS} in 2000 as a generalization of a metric
space. They generalized the Banach Contraction Principle \cite{Ba} in
such spaces.  Metric-like spaces were discovered by Amini-Harandi \cite{Amini}
who established some fixed point results.  Very recently, Karapinar and
Salimi \cite{Erdal1} established some fixed point theorems for  cyclic
contractions in the setting of metric-like spaces.
Many other (common) fixed point results in the context of metric-like
(quasi) spaces have been proved, see for example
\cite{Aage,Aage2,Erdal1,Koh,Ren,Zey,Zota}.

In the sequel, the letters $\mathbb{R}$, $\mathbb{R}^+_0$ and $\mathbb{N}^{\ast }$
will denote the set of real numbers, the set of nonnegative real numbers
and the set of positive integer numbers, respectively.


\begin{definition}[\cite{Amini}] \label{def1.1} \rm
Let $X$ be a nonempty set. A function $\sigma:X\times X\to \mathbb{R}^+_0$
is said to be a dislocated (metric-like) metric on $X$ if for any
$x,y,z\in X$, the following conditions hold:
\begin{itemize}
\item[(S1)] $\sigma(x,y)=\sigma(x,x)=0\Longrightarrow x=y$;

\item[(S2)] $\sigma(x,y)=\sigma(y,x)$;

\item[(S3)] $\sigma(x,z)\leq \sigma(x,y)+\sigma(y,z)$.

\end{itemize}
The pair $(X,\sigma)$ is then called a dislocated (metric-like) metric space.
\end{definition}

\begin{example}  \rm
A trivial example of a  metric-like space is the pair $(\mathbb{R}^+_0,\sigma)$,
where $\sigma:\mathbb{R}^+_0\times \mathbb{R}^+_0\to \mathbb{R}^+_0$ is
defined as $\sigma(x,y)=\max \{x,y\}$. Here, $\sigma$ is also a partial metric \cite{Ma}.
\label{Ex1}
\end{example}

\begin{example} \label{examp13} \rm
Take $X=\mathbb{R}$ and define the $\sigma$ metric-like as
\[
\sigma(x,y)=\frac{|x-y|+|x|+|y|}{2}\quad\text{for all }x,y\in X.
\]
Notice that $\sigma$ is not a metric. Particularly, if $X=\mathbb{R}^+_0$,
we have $\sigma(x,y)=\max\{x,y\}$ and so we return to Example \ref{Ex1}.
But, if $X=\mathbb{R}$, we have $\sigma(x,y)\neq \max\{x,y\}$.
\end{example}

As it is well  known, a partial metric \cite{Ma} is a metric-like.
The converse is not true. The following example concerns this  statement.

\begin{example} \label{examp1.4} \rm
Take  $X=\{1,2,3\}$ and consider the  metric-like
$\sigma:X \times X\to  \mathbb{R}^+_0$ given by
\begin{gather*}
\sigma(1,1)=0,\quad \sigma(2,2)=1,\quad
\sigma(3,3)=\frac{2}{3}, \quad
\sigma(1,2)=\sigma(2,1)=\frac{9}{10},\\
 \sigma(2,3)=\sigma(3,2)=\frac{4}{5},\quad
\sigma(1,3)=\sigma(3,1)=\frac{7}{10}.
\end{gather*}
Since $\sigma(2,2)\neq 0$, so $\sigma$ is not a metric and  since $\sigma(2,2)>\sigma(1,2)$,
so $\sigma$ is not a partial metric.
\end{example}

Each metric-like $\sigma$ on $X$ generates a $T_{0}$ topology $\tau _{\sigma}$ on
$X $ which has as a base the family open $\sigma$-balls
$\{B_{\sigma}(x,\varepsilon ):x\in X,\varepsilon >0\}$, where
$B_{\sigma}(x,\varepsilon )=\{y\in X:|\sigma(x,y)-\sigma(x,x)|<\varepsilon \}$,
 for all $x\in X$ and $\varepsilon >0$.

Observe  that a sequence $\{x_{n}\}$ in a metric-like space $(X,\sigma)$
converges to a point $x\in X$, with respect to $\tau _{\sigma}$, if and only
if $\sigma(x,x)=\lim_{n\to \infty }\sigma(x,x_{n})$.


\begin{definition}[\cite{Amini}] \rm
Let $(X,\sigma)$ be a metric-like space.
\begin{itemize}
\item[(a)] A sequence $\{x_{n}\}$ in $X$ is a Cauchy sequence if $
\lim_{n,m\to \infty }\sigma(x_{n},x_{m})$ exists and is finite.

\item[(b)] $(X,\sigma)$ is  complete if every Cauchy sequence
$\{x_{n}\} $ in $X$ converges with respect to $\tau _{\sigma}$ to a point
$x\in X$; that is,
\[
\lim_{n\to \infty }\sigma(x,x_{n})=\sigma(x,x)
=\lim_{n,m\to \infty }\sigma(x_n,x_{m}).
\]
\end{itemize}
\end{definition}

\begin{definition}[\cite{Amini}] \rm
Let $(X,\sigma)$ be a metric-like space. A mapping $T:(X,\sigma)\to (X,\sigma)$
is continuous if for any sequence $\{x_n\}$  in $X$ such that
 $\sigma(x_n,x)\to \sigma(x,x)$ as $n\to \infty$, we have $\sigma(Tx_n,Tx)\to \sigma(Tx,Tx)$
as $n\to \infty$.
\end{definition}

\begin{lemma}[\cite{Erdal1}] \label{L1}
Let $(X,\sigma)$ be a metric-like space. Let $\{x_n\}$ be a sequence in
 $X$ such that $x_n\to x$ where $x\in X$ and $\sigma(x,x)=0$. Then, for
all $y\in X$, we have
\[
\lim_{n\to \infty} \sigma(x_n,y)=\sigma(x,y).
\]
\end{lemma}

Let $\Psi $ be the family of functions $\psi :[0,\infty )\to [0,\infty )$
satisfying the following conditions:
\begin{itemize}
\item[(i)] $\psi $ is nondecreasing;

\item[(ii)] $\sum_{n=1}^{+\infty} \psi ^{n}(t)< \infty$ for all $t>0$.
\end{itemize}
Note that if $\psi\in\Psi$, we have $\psi(t)<t$ for all $t>0$.

In 2012, Samet et al \cite{Samet1} introduced the class of $\alpha$-admissible
mappings.

\begin{definition} \cite{Samet1}
For a nonempty set  $X$, let $T: X\to X$ and $\alpha: X\times X\to [0,\infty)$
be given mappings. We say that $T$ is $\alpha$-admissible  if  for all
 $x,y\in X$, we have
  \begin{equation}  \label{eqA}
  \alpha(x,y)\geq 1 \Longrightarrow \alpha(Tx,Ty)\geq 1.
  \end{equation}
  \end{definition}

The notion of  $\alpha-\psi$-contractive mappings is also defined
in the following way.

\begin{definition}[\cite{Samet1}] \rm
Let $(X,d)$ be a metric space and $T: X\to X$ be a given mapping.
We say that $T$ is a $\alpha-\psi$ contractive mapping if there exist two functions
$\alpha: X\times X \to [0, \infty)$ and  $\psi\in\Psi$ such that
 \begin{equation}  \label{eq1x}
\alpha(x,y)d(Tx,Ty)\leq\psi(d(x,y)), \text{ for all } x,y\in X.
  \end{equation}
\end{definition}

Many authors have proved fixed point results for generalized contractions
using the function $\alpha$, see for instance \cite{AK,AJK,AJK2,KS}.
Now, we state in the following definition a generalization  of
the notion of $\alpha-\psi$ contractive mappings in the context of a
metric-like space.

\begin{definition} \rm
Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping.
We say that $T$ is a generalized $\alpha-\psi$ contractive mapping of
type A if there exist two functions
$\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that
 \begin{equation}  \label{e1g}
\alpha(x,y)\sigma(Tx,Ty)\leq\psi(M(x,y)), \quad \text{ for all } x,y\in X,
  \end{equation}
  where
   \begin{equation}  \label{Mxy}
M(x,y)=\max\{\sigma(x,y),\sigma(x,Tx),\sigma(y,Ty),\frac{\sigma(x,Ty)+\sigma(y,Tx)}{4}\}.
  \end{equation}
\end{definition}

Our aim in this article is to provide some fixed point results for variant
generalized $\alpha-\psi$ contractive mappings in the setting of metric-like spaces.
We support our obtained theorems by some concrete examples and an application.

\section{Main results}

Our first fixed point result read as follows.

\begin{theorem} \label{thm1}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be a generalized
$\alpha-\psi$ contractive mapping of type A.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] $T$ is continuous.
\end{itemize}
Then  there exists a  $u\in X$ such that $\sigma(u,u)=0$. Assume in addition that
\begin{itemize}
\item[(H1)]
 If $\sigma(x,x)=0$ for some $x\in X$, then $\alpha(x,x)\geq 1$.
\end{itemize}
Then such $u$ is a fixed point of $T$, that is, $Tu=u$.
\end{theorem}

\begin{proof}
By assumption (ii), there exists a point  $x_0\in X$   such that
$\alpha(x_{0},Tx_{0})\geq 1$.
We define a sequence  $\{x_n\}$ in $X$ by $x_{n+1}=Tx_{n}=T^{n+1} x_0$
for all $n\geq 0$.
Suppose that $x_{n_0}=x_{n_0+1}$ for some $n_0$. So the proof is completed
since $u=x_{n_0}=x_{n_0+1}=Tx_{n_0}=Tu$.
Consequently, throughout the proof, we  assume that
\begin{equation}\label{e2a}
x_{n}\neq x_{n+1} \quad \text{for all }  n.
\end{equation}
Observe that
\[
 \alpha(x_{0},x_{1})=\alpha(x_{0},Tx_{0})\geq 1\Rightarrow
\alpha(Tx_{0},Tx_{1})=\alpha(x_{1},x_{2})\geq 1,
 \]
since $T$ is $\alpha$-admissible. By repeating the process above, we derive that
\begin{equation}\label{e2}
\alpha(x_{n},x_{n+1})\geq 1, \quad \text{for all } n=0,1,\ldots.
\end{equation}
\smallskip

\noindent\textbf{Step 1:} We shall prove that
\begin{equation}\label{e7EKC1}
 \lim_{n\to \infty}  \sigma(x_{n},x_{n+1})=0.
\end{equation}
Combining  \eqref{e1g} and \ref{e2}, we find that
\begin{equation}\label{e3C1}
\sigma(x_{n},x_{n+1})=\sigma(Tx_{n-1},Tx_{n})
\leq \alpha(x_{n-1},x_{n})\sigma(Tx_{n-1},Tx_{n})\leq\psi(M(x_{n-1},x_{n})),
\end{equation}
for all $n\geq 1$,
where
\begin{equation}\label{g1}
\begin{aligned}
&M(x_{n-1},x_{n})\\
&=\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n-1},Tx_{n-1}),\sigma(x_{n},Tx_{n}),
 \frac{\sigma(x_{n-1},Tx_{n})+\sigma(x_{n},Tx_{n-1})}{4}\}\\
&\\
&=\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1}),
 \frac{\sigma(x_{n-1},x_{n+1})+\sigma(x_{n},x_{n})}{4}\}\\
&\\
&\leq \max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1}),
 \frac{\sigma(x_{n-1},x_{n})+3\sigma(x_{n},x_{n+1})}{4}\}\\
&= \max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\}.
\end{aligned}
\end{equation}
If for some $n$,  $\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\}
=\sigma(x_{n},x_{n+1}) (\neq 0)$, then \eqref{e3C1} and \eqref{g1} turn into
\[
\sigma(x_{n},x_{n+1})\leq\psi(M(x_{n-1},x_{n}))
\leq \psi(\sigma(x_{n},x_{n+1}))<\sigma(x_{n},x_{n+1}),
 \]
which is a contradiction. Hence,
$\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\}=\sigma(x_{n-1},x_{n})$ for all
$n\in \mathbb{N}^*$ and \eqref{e3C1} becomes
 \begin{equation}\label{e3C1+}
\sigma(x_{n},x_{n+1})\leq\psi(\sigma(x_{n-1},x_{n}))\quad\text{for all } n\geq 1.
\end{equation}
This yields
\begin{equation} \label{aa1}
\sigma(x_{n},x_{n+1})<\sigma(x_{n-1},x_{n})\quad\text{for all } n\geq 1.
\end{equation}
By \eqref{e3C1+}, we find  that
 \begin{equation}\label{e7C1}
\sigma(x_{n},x_{n+1})\leq \psi^n (\sigma(x_0,x_{1})), \quad\text{for all } n\in \mathbb{N}.
\end{equation}
By the properties of $\psi$, we have
 \[
 \lim_{n\to \infty}  \sigma(x_{n},x_{n+1})=0.
\]
\smallskip

\noindent\textbf{Step 2:} We shall prove that $\{x_n\}$ is a Cauchy sequence.
First, by using (S3) and \eqref{e7C1}
\begin{equation}\label{C1EK}
\begin{aligned}
\sigma(x_n, x_{n+k})
&\leq  \sigma(x_{n},x_{n+1})+ \sigma(x_{n+1},x_{n+2})+\ldots+ \sigma(x_{n+k-1},x_{n+k})\\
&\leq  \sum_{p=n}^{n+k-1}\psi^p (\sigma(x_0,x_{1}))\\
&\leq  \sum_{p=n}^{+\infty}\psi^p (\sigma(x_0,x_{1})) \to 0 \quad \text{as } n\to \infty.
\end{aligned}
\end{equation}
Thus, by  the symmetry of $\sigma$, we obtain
\begin{equation}
\label{b9}
\lim_{n,m\to \infty} \sigma(x_n, x_{m})=0.
\end{equation}
We conclude that  $\{x_n\}$ is a Cauchy sequence in $(X,\sigma)$.
Since $(X,\sigma)$ is complete, there exists $u\in X$ such that
\begin{equation}\label{e8}
 \lim_{n\to \infty} \sigma(x_n,u)=\sigma(u,u)= \lim_{n,m\to \infty} \sigma(x_n,x_m)=0.
\end{equation}
Since $T$ is continuous, from \eqref{e8} we obtain that
\begin{equation}\label{e9}
 \lim_{n\to \infty} \sigma(x_{n+1},Tu)=\lim_{n\to \infty} \sigma(Tx_{n},Tu)=\sigma(Tu,Tu),
\end{equation}
On the other hand, by \eqref{e8} and Lemma \ref{L1}
 \begin{equation}\label{e10}
 \lim_{n\to \infty} \sigma(x_{n+1},Tu)=\sigma(u,Tu).
\end{equation}
Comparing \eqref{e9} and \eqref{e10}, we get $\sigma(u,Tu)=\sigma(Tu,Tu)$. By \eqref{e1g},
\[
\alpha(u,u)\sigma(Tu,Tu)\leq\psi(M(u,u)),
  \]
where
\begin{align*}
M(u,u)
&=\max\{\sigma(u,u),\sigma(u,Tu),\sigma(u,Tu),\frac{\sigma(u,Tu)+\sigma(u,Tu)}{4}\}\\
&=\max\{0,\sigma(u,Tu)\}=\sigma(u,Tu).
\end{align*}
 From hypothesis (H1) and the fact that $\sigma(u,u)=0$, we have
$\alpha(u,u)\geq 1$. Therefore, by \eqref{e1g}
  \[
  \sigma(u,Tu)\leq \alpha(u,u)\sigma(u,Tu)\leq \psi(\sigma(u,Tu)),
  \]
  which holds unless $\sigma(u,Tu)=0$, that is $Tu=u$. So $u$ is a fixed point of $T$.
\end{proof}


Theorem \ref{thm1} remains true if we replace the continuity hypothesis
by the following property:
\begin{quote}
If $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$
for all $n$ and $x_n \to x\in X$ as $n\to \infty$,
then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that
$\alpha(x_{n(k)},x)\geq 1$ for all $k$.
\end{quote}
This statement is given as follows.

\begin{theorem}\label{thm2}
Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized
$\alpha-\psi$ contractive mapping of type A.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that
$\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as
$n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$
such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$.
\end{itemize}
Then, there exists   $u\in X$ such that $Tu=u$.
\end{theorem}

\begin{proof}
Following the proof of Theorem \ref{thm1}, we know that the sequence
$\{x_n\}$ defined by $x_{n+1}=Tx_n$ for all $n\geq 0$ is Cauchy in $(X,\sigma)$
and converges to some $u\in X$. Also, \eqref{e8}) holds, so
\begin{equation}
\lim_{k\to \infty} \sigma(x_{n(k)+1},Tu)=\sigma(u,Tu).
\label{b10}
\end{equation}
We shall show that $Tu=u$. Suppose, on the contrary, that  $Tu\neq u$,
i.e, $\sigma(Tu,u)>0$.
From \eqref{e2} and condition (iii), there exists a subsequence
$\{x_{n(k)}\}$ of $\{x_n\}$ such that  $\alpha(x_{n(k)},u)\geq 1$ for all $k$.

 By applying \eqref{e1g},  we obtain
\begin{equation}\label{h1}
 \sigma(x_{n(k)+1},Tu)\leq \alpha(x_{n(k)},u)\sigma(Tx_{n(k)}, Tu)\leq \psi(M(x_{n(k)},u))
\end{equation}
where
\begin{equation}\label{10aks}
\begin{split}
&M(x_{n(k)},u)\\
&=\max\{\sigma(x_{n(k)},u),\sigma(x_{n(k)},Tx_{n(k)}),\sigma(u,Tu),
 \frac{\sigma(x_{n(k)},Tu)+\sigma(u,Tx_{n(k)})}{4}\}\\
&=\max\{\sigma(x_{n(k)},u),\sigma(x_{n(k)},x_{n(k)+1}),\sigma(u,Tu),
 \frac{\sigma(x_{n(k)},Tu)+\sigma(u,x_{n(k)+1})}{4}\}.
\end{split}
\end{equation}
By \eqref{e7EKC1} and \eqref{b10}, we have
\begin{equation}
\lim_{k\to \infty} M (x_{n(k)},u)=\sigma(u,Tu).
\label{b11}
\end{equation}
Letting $k\to \infty$ in \eqref{h1})
\begin{equation}\label{e11}
\sigma(u,Tu) \leq \psi(\sigma(u,Tu)) <\sigma(u,Tu),
\end{equation}
which is a contradiction. Hence,  we obtain that  $u$ is a fixed point
of $T$, that is, $Tu=u$.
\end{proof}

\begin{definition} \rm
Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping.
We say that $T$ is a  generalized  $\alpha-\psi$ contractive mapping 
of  type B if there exist two functions
$\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that
 \begin{equation}  \label{e1gN}
\alpha(x,y)\sigma(Tx,Ty)\leq\psi(M_0(x,y)), \text{ for all } \,x,y\in X,
  \end{equation}
  where
   \begin{equation}  \label{Nxy}
M_0(x,y)=\max\{\sigma(x,y),\sigma(x,Tx),\sigma(y,Ty)\}.
  \end{equation}
\end{definition}

\begin{theorem}\label{thm2c}
Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized 
$\alpha-\psi$ contractive mapping of type B.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] $T$ is continuous.
\end{itemize}
Then  there exists a  $u\in X$ such that $\sigma(u,u)=0$. 
If in addition {\rm (H1)} holds,
then such $u$ is a fixed point of $T$, that is, $Tu=u$.
\end{theorem}

\begin{proof}
Along the lines of the proof of Theorem \ref{thm1}, we get the desired result.
Because of the analogy, we skip the details of the proof.
\end{proof}

\begin{theorem}\label{thm2d}
Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized 
$\alpha-\psi$ contractive mapping of type B.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$
 for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence 
$\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$.
\end{itemize}
Then, there exists   $u\in X$ such that $Tu=u$.
\end{theorem}

We omit the proof because of the similarity to  Theorem  \ref{thm2}.

\begin{definition} \rm
Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping.
We say that $T$ is a  $\alpha-\psi$ contractive mapping  if there exist 
two functions $\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that
 \begin{equation}  \label{e1gNg}
\alpha(x,y)\sigma(Tx,Ty)\leq\psi(\sigma(x,y)), \quad \text{for all } x,y\in X,
  \end{equation}
\end{definition}

\begin{theorem}\label{thm2e}
Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a   
$\alpha-\psi$ contractive mapping.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] $T$ is continuous.
\end{itemize}
Then  there exists a  $u\in X$ such that $\sigma(u,u)=0$. 
If in addition {\rm (H1)} holds, 
then such $u$ is a fixed point of $T$, that is, $Tu=u$.
\end{theorem}

The above theorem is a simple consequence of Theorem \ref{thm1}.

\begin{theorem}\label{thm2f}
Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a 
 $\alpha-\psi$ contractive mapping.
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that 
$\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as 
$n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of 
$\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$.
\end{itemize}
Then, there exists   $u\in X$ such that $Tu=u$.
\end{theorem}

The above theorem follows from Theorem  \ref{thm2}.


\section{Consequences of the main results}
In the following, we present some illustrated consequences of our obtained 
results  given by Theorem \ref{thm1} and Theorem \ref{thm2}.

\subsection{Standard fixed point results in metric-like spaces}
\label{con1}

\begin{corollary} \label{Cor1}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \psi(M(x,y))\quad\text{for all }x,y\in X
\]
where $M(x,y)$ is defined by \eqref{Mxy}.  Then, $T$ has a fixed point.
\end{corollary}

To prove the above corollary it suffices to take $\alpha(x,y)=1$ in 
Theorem \ref{thm2}.

\begin{corollary}\label{Cor2}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \lambda\,M(x,y)\quad\text{for all}\,\,x,y\in X
\]
where $\lambda\in [0,1)$.  Then, $T$ has a fixed point.
\end{corollary}

\begin{proof}
To prove the above corollary it suffices to take $\psi(t)=\lambda t$ 
in Corollary \ref{Cor1}.
\end{proof}

\begin{corollary} \label{Cor1b}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \psi(M_0(x,y))\quad\text{for all}\,\,x,y\in X
\]
where $M_0(x,y)$ is defined by \eqref{Nxy}.  Then, $T$ has a fixed point.
\end{corollary}

To prove the above corollary it
 suffices to take $\alpha(x,y)=1$ in Theorem \ref{thm2c}.

\begin{corollary}\label{Cor2b}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \lambda\,M_0(x,y)\quad\text{for all}\,\,x,y\in X
\]
where $\lambda\in [0,1)$.  Then, $T$ has a fixed point.
\end{corollary}

To prove the above corollary it suffices to take $\psi(t)=\lambda t$ 
in Corollary \ref{Cor1b}.

\begin{corollary}\label{Cor1c}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \psi(\sigma(x,y))\quad\text{for all}\,\,x,y\in X.
\]
 Then, $T$ has a fixed point.
\end{corollary}

To prove the above corollary it suffices to take $\alpha(x,y)=1$ 
in Theorem \ref{thm2f}.

\begin{corollary}\label{Cor2c}
Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that
\[
\sigma(Tx,Ty)\leq \lambda\,\sigma(x,y)\quad\text{for all}\,\,x,y\in X
\]
where $\lambda\in [0,1)$.  Then, $T$ has a fixed point.
\end{corollary}

To prove the above corollary it suffices to take $\psi(t)=\lambda t$ in 
Corollary \ref{Cor1c}.

\subsection{Standard fixed point results in partial metric spaces}
\label{con2}
The partial metric spaces were introduced
by Matthews \cite{Ma} as a part of the study of denotational semantics 
of data for networks.

\begin{definition}[\cite{Ma}] \label{partial}\rm
 A partial metric on a nonempty set $X$ is a function $p :
X\times X\to [0, +\infty)$ such that for all $x, y, z \in X$:
\begin{itemize}
\item[(P1)] $x =y\Longleftrightarrow p(x,x)=p(x,y)= p(y,y)$,

\item[(P2)] $p(x,x) \leq p(x,y)$,

\item[(P3)] $p(x,y)= p(y,x)$,

\item[(P4)] $p(x,y)\leq p(x,z)+ p(z,y)-p(z,z)$.

\end{itemize} 
A partial metric space is a pair $(X,p)$ such that $X$ is a nonempty set and
$p$ is a partial metric on $X$.
\end{definition}

If $p$ is a partial metric on $X$, then the function $d_p : X\times X
\to \mathbb{R}^+_0$ given by
\begin{equation}  \label{ps}
d_p(x,y)=2p(x,y)-p(x,x)-p(y,y),
\end{equation}
is a metric on $X$.

\begin{lemma} \label{le-s}
 Let $(X,p)$ be a partial metric space. Then,  (a) $\{x_{n}\}$
is a Cauchy sequence in $(X,p)$ if and only if it is a Cauchy sequence in
the metric space $(X,d_p)$, (b) $X$ is complete if and only if the metric
space $(X,d_p)$ is complete.
\end{lemma}

\begin{corollary} \label{Cor3}
Let $(X,\sigma)$ be a complete partial space and $T: (X,p)\to (X,p)$ be such that
\[
\alpha(x,y) p(Tx,Ty)\leq \psi(N(x,y))\quad\text{for all}\,\,x,y\in X
\]
where $N(x,y)$ is defined as
\[
N(x,y)=\max\{p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2}\}.
\]
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] $T$ is continuous.
\end{itemize}
Then  there exists a  $u\in X$ such that $p(u,u)=0$. If in addition
{\rm (H1)} holds, 
then such $u$ is a fixed point of $T$, that is, $Tu=u$.
\end{corollary}

\begin{proof} It  suffices to replace the metric-like $\sigma$ 
in Theorem \ref{thm1} by the partial metric $p$ which itself a metric-like. 
Note that we considered in $N(x,y)$ the fourth term $\frac{p(x,Ty)+p(y,Tx)}{2}$ 
instead of $\frac{p(x,Ty)+p(y,Tx)}{4}$ due to the inequality $p(x,x)\leq p(x,y)$.
 Its proof is evident.
\end{proof}

Similar to Corollary \ref{Cor3},  from Theorem \ref{thm2} we  deduce the following
result.

\begin{corollary} \label{Cor4}
Let $(X,\sigma)$ be a complete partial space and $T: (X,p)\to (X,p)$ be such that
\[
\alpha(x,y) p(Tx,Ty)\leq \psi(N(x,y))\quad\text{for all}\,\,x,y\in X.
\]
 Suppose  that
\begin{itemize}
\item[(i)] $T$ is $\alpha$-admissible;
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$;
\item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that 
$\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as 
$n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ 
such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$.
\end{itemize}
Then, there exists   $u\in X$ such that $Tu=u$.
\end{corollary}

\begin{remark} \label{rmk3.11} \rm
It is clear that one can easily state the analog of Theorem \ref{thm2c},  
Theorem \ref{thm2d},  Theorem \ref{thm2e} and
Theorem \ref{thm2f} in the setting of partial metric spaces.
\end{remark}

\subsection{Fixed point results with a partial order}
\label{con3}
The study of the existence of fixed points on metric spaces endowed with 
a partial order can be considered as
one of the very interesting improvements in the field  of fixed point theory. 
 This trend was initiated by Turinici \cite{T} in 1986, but it became one of 
the core research subject after the publications of  Ran and Reurings in \cite{RR}
and  Nieto and  Rodr\'iguez-L\'opez \cite{NR}.

\begin{definition} \rm
Let $(X,\preceq)$ be a partially ordered set and $T: X\to X$ be a given mapping.
 We say that $T$ is nondecreasing with respect to $\preceq$ if
$$
x,y\in X,\,\,x\preceq y \Longrightarrow Tx\preceq Ty.
$$
\end{definition}

\begin{definition} \rm
 Let $(X,\preceq)$ be a partially ordered set. A sequence $\{x_n\}\subset X$ 
is said to be nondecreasing with respect to
$\preceq$ if $x_n\preceq x_{n+1}$ for all $n$.
\end{definition}

\begin{definition}
Let $(X,\preceq)$ be a partially ordered set and $\sigma$ be a metric-like on $X$.
We say that $(X,\preceq,\sigma)$ is regular if for every nondecreasing sequence 
$\{x_n\}\subset X$ such that
$x_n\to x\in X$ as $n\to \infty$, there exists a subsequence 
$\{x_{n(k)}\}$ of $\{x_n\}$ such that
$x_{n(k)}\preceq x$ for all $k$.
\end{definition}

\begin{corollary}\label{CT7}
Let $(X,\preceq)$ be a partially ordered set and $\sigma$ be a metric-like on $X$ 
such that $(X,\sigma)$ is complete.  Let $T: X\to X$ be a nondecreasing mapping 
with respect to $\preceq$. Suppose that there exists a function $\psi\in \Psi$ 
such that
$$
\sigma(Tx,Ty)\leq \psi(M(x,y)),
$$
for all $x,y\in X$ with $x\succeq y$. Suppose also that the following conditions 
hold:
\begin{itemize}
\item[(i)] there exists $x_0\in X$ such that $x_0\preceq Tx_0$;
\item[(ii)] ($T$ is continuous and the property $(H)$  holds) or ($(X,\preceq,\sigma)$ is regular).
\end{itemize}
Then, $T$ has a fixed point.
\end{corollary}

\begin{proof}
Define the mapping $\alpha: X\times X\to [0,\infty)$ by
\[
\alpha(x,y)=\begin{cases}
1 &\text{if }  x\preceq y \text{ or } x\succeq y,\\
0 &\text{otherwise}.
\end{cases}
\]
Clearly, $T$ is a generalized $\alpha-\psi$ contractive mapping of 
type A; that is,
$$
\alpha(x,y)\sigma(Tx,Ty)\leq \psi(M(x,y)),
$$
for all $x,y\in X$. From condition (i), we have $\alpha(x_0,Tx_0)\geq 1$. 
 Moreover, for all $x,y\in X$, from the monotone property of $T$, we have
$$
\alpha(x,y)\geq 1 \Longrightarrow x\succeq y \text{ or }
 x\preceq y  \Longrightarrow Tx\succeq Ty \text{ or } 
Tx\preceq Ty \Longrightarrow \alpha(Tx,Ty)\geq 1.
$$
Thus, $T$ is $\alpha$-admissible. Now, if $T$ is continuous and the hypothesis 
(H1) holds, the existence of a fixed point follows from Theorem \ref{thm1}.
Suppose now that $(X,\preceq,d)$ is regular. Let $\{x_n\}$ be a sequence in 
$X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ 
as $n\to \infty$. From the regularity hypothesis, there exists a subsequence 
$\{x_{n(k)}\}$ of $\{x_n\}$ such that
$x_{n(k)}\preceq x$ for all $k$. This implies from the definition of 
$\alpha$ that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. 
In this case, the existence of a fixed point follows from Theorem \ref{thm2}.
\end{proof}

\begin{remark} \label{rmk} \rm
Notice that we may obtain the analog of Theorem \ref{thm2c},  
Theorem \ref{thm2d},  Theorem \ref{thm2e},
Theorem \ref{thm2f} and the results of Subsection \ref{con1} 
and Subsection \ref{con2}  in the setting of partially ordered metric-like spaces.
\end{remark}


\subsection{Fixed point results for cyclic contractions}

 Kirk, Srinivasan and Veeramani \cite{Kirk} proved very interesting
 generalizations of the Banach Contraction Mapping Principle by introducing a 
cyclic contraction.  This remarkable paper \cite{Kirk}  has been appreciated 
 by many  several researchers (see, for example, \cite{CC3, CC4, CC3P, Petric, CC2} 
and the related reference therein).
In this subsection, we  derive some fixed point theorems for cyclic contractive 
mappings in the setting of metric-like spaces.

\begin{corollary}\label{CT14}
Let $ \{A_i\}_{i=1}^2$ be nonempty closed subsets of a complete metric-like 
space $(X,\sigma)$ and
$T:Y \to Y$ be a given mapping, where $Y=A_1\cup A_2$.  
Suppose that the following conditions hold:
\begin{itemize}
\item[(I)] $T(A_1)\subseteq A_2$ and $T(A_2)\subseteq A_1$;
\item[(II)] there exists a function $\psi\in \Psi$ such that
$$
\sigma(Tx,Ty)\leq \psi(M(x,y)), \text{ for all } (x,y)\in A_1\times A_2.
$$
\end{itemize}
Then $T$ has a fixed point that belongs to  $A_1\cap A_2$.
\end{corollary}



\begin{proof}
Since $A_1$ and $A_2$ are closed subsets of the complete metric-like space 
$(X,\sigma)$, then $(Y,\sigma)$ is complete.
Define the mapping $\alpha: Y\times Y\to [0,\infty)$ by
\[
\alpha(x,y)=\begin{cases}
1 &\text{if } (x,y)\in (A_1\times A_2)\cup   (A_2\times A_1),\\
0 &\text{otherwise.}
\end{cases}
\]
From (II) and the definition of $\alpha$, we can write
$$
\alpha(x,y)d(Tx,Ty)\leq \psi(M(x,y)),
$$
for all $x,y\in Y$. Thus $T$ is a generalized $\alpha-\psi$ contractive 
mapping of type A.
Let $(x,y) \in Y\times Y$ such that $\alpha(x,y)\geq 1$.

If $(x,y)\in A_1\times A_2$, from (I),
$(Tx,Ty)\in A_2\times A_1$, which implies that $\alpha(Tx,Ty)\geq 1$.

If $(x,y)\in A_2\times A_1$, from (I),
$(Tx,Ty)\in A_1\times A_2$, which implies that $\alpha(Tx,Ty)\geq 1$.


Hence,  in all cases, we conclude that $\alpha(Tx,Ty)\geq 1$ which yields that
that $T$ is $\alpha$-admissible.

Notice also that, from (I), for any $a\in A_1$, we have $(a,Ta)\in A_1\times A_2$,
which implies that $\alpha(a,Ta)\geq 1$.

Now, let $\{x_n\}$ be a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ 
for all $n$ and $x_n \to x\in X$ as $n\to \infty$. This implies from the 
definition of $\alpha$ that
$$
(x_n,x_{n+1})\in (A_1\times A_2)\cup   (A_2\times A_1), \quad\text{for all } n.
$$
Since $(A_1\times A_2)\cup   (A_2\times A_1)$ is a closed set with respect 
to the metric-like $\sigma$, we get that
$$
(x,x)\in (A_1\times A_2)\cup   (A_2\times A_1),
$$
which implies that $x\in A_1\cap A_2$. Thus we get immediately from the 
definition of $\alpha$ that
$\alpha(x_n,x)\geq 1$ for all $n$.

Now, all the hypotheses of  Theorem \ref{thm2} are satisfied and $T$ 
has a fixed point in $Y$.\bigskip
\end{proof}

Note that Corollary \ref{CT14} is a generalization of 
\cite[Corollary 1.10]{Erdal1}.


\section{Examples}

We present the following two concrete examples to support our results.

\begin{example} \label{examp4.1} \rm
Consider $X=\{0,1,2\}$. Take the metric-like $\sigma:X\times X\to \mathbb{R}^+_0$ 
defined by
\begin{gather*}
\sigma(0,0)=\sigma(1,1)=0,\quad \sigma(2,2)=\frac{9}{20}, \\
\sigma(0,2)=\sigma(2,0)=\frac{2}{5},\quad \sigma(1,2)=\sigma(2,1)=\frac{3}{5},\\
\sigma(0,1)=\sigma(1,0)=\frac{1}{2}.
\end{gather*}
Note that $\sigma(2,2)\neq 0$, so $\sigma$ is not a metric and $\sigma(2,2)>\sigma(0,2)$, 
so $\sigma$ is not a partial metric. Clearly, $(X,\sigma)$ is a complete 
metric-like space.
Given $T:X\to X$ as
$T0=T1=0$ and $T2=1$.
Take $\psi(t)=5t/6$ for each $t\geq 0$. Define the mapping $\alpha: X\times X\to [0,\infty)$ by
\[
\alpha(x,y)=\begin{cases}
1 & \text{if }  x=0,\\
0 & \text{otherwise}.
\end{cases}
\]
First, let $x,y\in X$ such that $\alpha(x,y)\geq 1$. By the definition 
of $\alpha$, this implies that $x=0$ and since $T0=0$, so $\alpha(Tx,Ty)=1$ 
for each $y\in X$, that is, $T$ is $\alpha$-admissible. We distinguish two cases:

\noindent Case 1: If ($x=0$ and $y=0$) or ($x=0$ and $y=1$), we have
\[
\alpha(Tx,Ty)\sigma(Tx,Ty)=\sigma(Tx,Ty)=0.
\]
Case 2: If $x=0$ and $y=2$, we have
\begin{align*}
 \alpha(Tx,Ty)\sigma(Tx,Ty)&=\sigma(Tx,Ty)=\sigma(0,1)=\frac{1}{2}=\frac{5}{6}\sigma(2,1)\\
 &=\psi(\sigma(y,Ty))\\
 &\leq \psi(M(x,y))
 \end{align*}
where $M(x,y)$ is defined by \eqref{Mxy}. It is also obvious that 
hypothesis (iii)  of Theorem \ref{thm2} is satisfied.
Thus, we map apply Theorem \ref{thm2} and so $T$ has a fixed point, 
which is $u=0$.
\end{example}


\begin{example} \label{examp4.2} \rm
Let $X=[0,\infty)$ be endowed with the metric-like $\sigma$ given as 
$\sigma(x,y)=\max\{x,y\}$. Define the mapping $T:X\to X$ by
\begin{equation*}
Tx=\begin{cases}
\frac{1}{2} x^2 & \text{if }x\in [0,1], \\
3x-1  & \text{otherwise}.
\end{cases}
\end{equation*}
Consider $\psi:[0,\infty)\to [0,\infty)$ defined by
\begin{equation*}
\psi(t)=\begin{cases}
\frac{1}{2} t^2 & \text{if } 0\leq t<1, \\
\frac{1}{2}   & \text{otherwise}.
\end{cases}
\end{equation*}
Obviously, $\psi\in \Psi$. Consider $\alpha: X\times X\to [0,\infty)$ as
\[
\alpha(x,y)=\begin{cases}
1 &\text{if }  x,y\in [0,1],\\
0 &\text{otherwise}.
\end{cases}
\]
First, let $x,y\in X$ such that $\alpha(x,y)\geq 1$, so $x,y\in [0,1]$. 
In this case,
\[
\alpha(Tx,Ty)=\alpha(\frac{1}{2} x^2,\frac{1}{2} y^2)=1;
\]
that is, $T$ is $\alpha$-admissible. Here we also have
\begin{align*}
\alpha(Tx,Ty)\sigma(Tx,Ty)
&=\sigma(Tx,Ty)=\sigma(\frac{1}{2} x^2,\frac{1}{2} y^2)\\
&=\sigma(\psi(x),\psi(y))=\max(\psi(x),\psi(y))\\
&=\psi(\max(x,y))=\psi(\sigma(x,y))\\
&\leq \psi(M(x,y)).
\end{align*}
Note that hypothesis (iii)  of Theorem \ref{thm2} is also satisfied. 
Applying Theorem \ref{thm2}, $T$ has a fixed point in $X$, which is $u=0$.
\end{example}


\section{Applications}

Here, we consider the following two-point boundary-value problem for the
second-order differential equation
\begin{equation} \label{ap1}
\begin{gathered}
-\frac{d^2x}{dt^2}=f(t,x(t)),\quad t\in [0,1]\\
x(0)=x(1)=0,
\end{gathered}
\end{equation}
where $f:[0,1]\times \mathbb{R}\to \mathbb{R}$  is a continuous function. 
Recall that the Green's function
associated to \eqref{ap1} is 
\begin{equation}\label{ap2}
G(t,s)= \begin{cases}
t(1-s) &\text{if } 0\leq t\leq s\leq1\\
s(1-t) &\text{if } 0\leq s\leq t\leq1.
\end{cases}
\end{equation}
Let $X=\mathcal{C}(I) (I = [0, 1])$  be the space of all continuous 
functions defined on $I$. We consider on $X$,  the metric-like $\sigma$  
given by
\[
\sigma(x,y)=\|x-y\|_{\infty}+\|x\|_{\infty}+\|y\|_{\infty}\quad\text{for all }
x,y\in X,
\]
where $\|u\|_{\infty}=\max_{t\in[0,1]} |u(t)|$ for each $u\in X$.

Note that $\sigma$ is also a partial metric on $X$ and since
\[
d_{\sigma} (x,y):=2\sigma(x,y)-\sigma(x,x)-\sigma(y,y)=2\|x-y\|_{\infty},
\]
so by Lemma \ref{le-s}, $(X,\sigma)$ is complete since the metric space
 $(X,\|\cdot\|_\infty)$ is complete.

It is well known that $x\in C^2(I)$ is a solution of \eqref{ap1}
is equivalent to that $x\in X=C(I)$ is a solution of the integral equation
\begin{equation}\label{ap4}
x(t) =\int_0^ 1 G(t, s) f (s, x(s)) ds, \quad\text{for all } t \in I.
\end{equation}


\begin{theorem}\label{t-ap}
Suppose the following conditions hold:\\
\begin{itemize}
\item there exists a continuous function $p: I\to \mathbb{R}^+_0$ such that
\[
|f (s, a)-f (s, b)|\leq 8\, p(s)\,|a-b|,
\]
for each $s\in I$ and $a,b\in \mathbb{R}$;

\item there exists a continuous function $q: I\to \mathbb{R}^+_0$ such that
\[
|f (s, a)|\leq 8\, q(s)\,|a|,
\]
for each $s\in I$ and $a\in \mathbb{R}$;

\item $\sup_{s\in I} p(s)=\lambda_1<\frac{1}{3}$;

\item $\sup_{s\in I} q(s)=\lambda_2<\frac{1}{3}$.

\end{itemize}
Then  problem \ref{ap1} has a solution $u\in X=\mathcal{C}(I,\mathbb{R})$.
\end{theorem}

\begin{proof}
 Consider the mapping $T:X\to X$ defined by
\[
Tx(t)=\int_0^ 1 G(t, s) f (s, x(s)) ds.
\]
for all  $x\in X$ and $t\in I$. Then, problem \eqref{ap1} is equivalent 
to finding $u\in X$ that is a fixed point of $T$.


Now, let $x,y\in X$. We have
\begin{align*}
 |Tx(t)-Ty(t)|
&=|\int_0^ 1 G(t, s) f (s, x(s)) ds-\int_0^ 1 G(t, s) f (s, y(s)) ds|\\
  &\leq \int_0^ 1 G(t, s) |f (s, x(s))-f (s, y(s))|\, ds\\
  &\leq 8\int_0^ 1 G(t, s) p(s)\,|x(s)-y(s)|\, ds\\
  &\leq 8\lambda_1 \|x-y\|_{\infty} \sup_{t\in I}\int_0^ 1 G(t, s)\,ds\\
  &= \lambda_1 \|x-y\|_{\infty}.
\end{align*}
In the above equality, we used  that for each $t\in I$, we have 
$\int_0^1 G(t,s)\,ds=-\frac{t^2}{2}+\frac{t}{2}$, and so 
$\sup_{t\in I}\int_0^ 1 G(t, s)\,ds=\frac{1}{8}$.
Therefore,
\begin{equation} \label{ap5}
\|Tx-Ty\|_{\infty}\leq  \lambda_1 \|x-y\|_{\infty}.
\end{equation}
Again, we have
\begin{align*}
 |Tx(t)|&=|\int_0^ 1 G(t, s) f (s, x(s)) ds|\\
  &\leq \int_0^ 1 G(t, s)\,|f (s, x(s))|\, ds\\
  &\leq 8\int_0^ 1 G(t, s)\, q(s)\,|x(s)|\, ds\\
  &\leq 8\lambda_2 \|x\|_{\infty} \sup_{t\in I}\int_0^ 1 G(t, s)\,ds\\
  &\leq \lambda_2 \|x\|_{\infty}.
\end{align*}
Thus
\begin{equation} \label{ap6}
\|Tx\|_{\infty}\leq  \lambda_2 \|x\|_{\infty}.
\end{equation}
Proceeding similarly, 
\begin{equation}\label{ap7}
\|Ty\|_{\infty}\leq  \lambda_2 \|y\|_{\infty}.
\end{equation}
Take $\lambda=\lambda_1+2\lambda_2$.  Under assumptions in Theorem  \ref{t-ap}, 
we have $\lambda<1$.
Summing \eqref{ap5} to \eqref{ap7}, we find
\begin{align*}
\sigma(Tx,Ty)
&=\|Tx-Ty\|_{\infty}+\|Tx\|_{\infty}+\|Ty\|_{\infty}\\
&\leq  \lambda_1 \|x-y\|_{\infty}+\lambda_2 \|x\|_{\infty}
+\lambda_2 \|y\|_{\infty}\\
&\leq (\lambda_1+2\lambda_2)(\|x-y\|_{\infty}+\|x\|_{\infty}+\|y\|_{\infty})\\
&= \lambda \sigma(x,y)\leq \lambda M(x,y).
\end{align*}
So all hypotheses of Corollary \ref{Cor2} are satisfied, and so $T$ has 
a fixed point $u\in X$, that is, the problem \eqref{ap1} has a 
solution $u\in C^2(I)$.
\end{proof}


\subsection*{Conclusion}
All fixed point results presented in this article are also valid 
for metric spaces. Consequently,
our results  extend and unify several results from the literature.

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