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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 181, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/181\hfil Continuous algebraic difference equations]
{Periodic solutions and asymptotic behavior for continuous
algebraic difference equations}

\author[E. Ait Dads, L. Lhachimi \hfil EJDE-2017/181\hfilneg]
{El Hadi Ait Dads, Lahcen Lhachimi}

\address{El Hadi Ait Dads \newline
Cadi Ayyad University,
Faculty of Sciences,
Department of Mathematics B.P. 2390,
 Marrakesh, Morocco}
\email{aitdads@uca.ac.ma}

\address{Lahcen Lhachimi \newline
 Unit\'e associ\'ee au CNRST (Morocco) URAC 02.\newline
UMI- UMMISCO (IRD- UPMC) France}
\email{lllahcen@gmail.com}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted March 22, 2017. Published July 14, 2017.}
\subjclass[2010]{39A13, 34C27}
\keywords{Bounded solution; periodic solution; asymptotic behavior;
\hfill\break\indent kernel theorem decomposition; Fourier coefficients}

\begin{abstract}
 Many phenomena in mathematical physics and in the theory of dynamical
 populations are described by difference equations. The aim of this work
 is to study existence of periodic solutions and the asymptotic behavior
 for some algebraic difference equations.
 The technique used is based on convergence of series associated with
 the forcing term and the characterization by Fourier coefficients.
 Our results generalize the main results of our previous results in \cite{al}.
 For illustration, we provide some examples.
\end{abstract}

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

\section{Introduction}

The existence problem of bounded solutions has been one of
the most attractive topics in the qualitative theory of ordinary or
functional differential equations for its significance in the physical
sciences. In many cases, it is of interest to model the evolution of some
system over time. There are two distinct cases. One can think of time as a
continuous variable, or one can think of time as a discrete variable. The
first case often leads to differential equations,
the second case leads to difference equations. We will not discuss
differential equations in this note.

Difference equations have many applications in population dynamics, they can
be used to describe the evolution of many phenomena over the course of
time. For example, if a certain population has discrete generations, the
size of the $(n+1)$th generation $x(n+1)$ is a function of the $n$th
generation $x(n)$. This relation express in the following difference
equation (see \cite{co})
\begin{equation}
x(n+1)=f(x(n)).  \label{1a}
\end{equation}

The main goal of this article is to study the problem of existence of periodic
solutions and their asymptotic behaviour of the  equation
\begin{equation}
x(t+1)-x(t)=f(t).  \label{a}
\end{equation}

Before, in its first part we consider the qualitative properties and
behavior at infinity of real continuous solutions of algebraic difference
equations of the  form
\begin{equation}
P(x(t+m),\dots ,x(t),t)=0.  \label{e}
\end{equation}

Where $P$ is a polynomial with real coefficients in its arguments
$x(t+m),\dots ,x(t)$ and $t$. The problem was first treated by Lancaster
\cite{lan}, who obtained an upper bound for the rate of increase of the solutions
of algebraic difference equations of a given order and pointed out the
surprising dissimilarity with the known rates of increase for solutions of
differential equations of the same order. The main goal of the second part
of this work is to investigate the problem of existence of periodic
solutions of the special difference equation \eqref{a}.
The work is motivated by some quantitative and qualitative results of the
difference equation considered in Ait Dads et al \cite{al}.

In the first section, we are concerned with the  algebraic
equation:
 Let $d\in \mathbb{N}^{\ast }$, $(a_{0},a_{1},..,a_{d-1})\in
\mathbb{C}^{d}$, with $a_{0}\neq 0$. Let us consider the following
difference equation:
\begin{equation}
f(t+d)=\underset{i=0}{\overset{d-1}{\sum }}a_if(t+i), \quad 
\forall t\in \mathbb{R},\label{aste}
\end{equation}
where $f\in C(\mathbb{R},\mathbb{C})$, for which we associate the polynomial
\[
P(X)=X^{d}-\underset{i=0}{\overset{d-1}{\sum }}a_iX^{i}
=\overset{r}{\underset{k=1}{\Pi }}(X-\lambda _k)^{m_k},\quad r\leq d
\]
with the $\lambda _k's$ are non null complexes which are two by
two distinct. And we have the first result concerning this equation.

The organization of the paper is as follows. Section $2$ concerns the study
of algebraic properties of some difference equation. In section $3$, we
consider  the asymptotic behavior for solutions of special difference
equation \eqref{a}. Section $4$ deals with the characterization of bounded
solutions of the equation \eqref{a}, this last section is divided into three
subsections, in the first one, we study the existence of periodic solutions
in connection with the forcing term of equation \eqref{a}, the second one is
concerned with the charaterization of periodic solution with respect to the
Fourier coefficients. The third subsection considers the case where the
period is an irrational number. In the last section, to illustrate the work,
some examples are given in the paper.

\section{Difference Equations}

\subsection{General solution of \eqref{aste}}\quad 

\begin{definition} \label{def1} \rm
The function $f$ is called general solution of equation \eqref{aste} if $f$
is continuous from $\mathbb{R}$ to $\mathbb{C}$ and satisfies
\eqref{aste}.
\end{definition}

\begin{proposition} \label{P1}
If we put $\lambda _k=e^{c_k}$. Then the solutions of \eqref{aste} are of the form:
\begin{equation*}
f(t)=\sum_{k=1}^r e^{c_kt}\sum_{i=0}^{m_k-1} p_i(t)t^{i}
\end{equation*}
where the $p_i's$ are continuous and $1$-periodic functions.
\end{proposition}

\begin{proof}
Let $E=C(\mathbb{R},\mathbb{C})$, and $\tau :E\to E$, defined by
\begin{equation*}
\tau (f)(t)=f(t+1), \forall f\in E.
\end{equation*}
Then the set of solutions of \eqref{aste} is $\ker P(\tau )$,
and by the Kernels decomposition theorem we have
\begin{equation*}
\ker P(\tau )=\overset{r}{\underset{k=1}{\oplus }}\ker (\tau -\lambda
_k\operatorname{id})^{m_k}.
\end{equation*}
Let $M_k:E\to E$ be defined for all $f\in E$ by $
M_k(f)(t)=e^{c_kt}f(t)$. One has
\begin{equation*}
\tau \circ M_k(f)(t)=e^{c_k(t+1)}f(t+1)=\lambda
_ke^{c_kt}f(t+1)=\lambda _kM_k\circ \tau (f)(t).
\end{equation*}
Then $\tau \circ M_k=\lambda _kM_k\circ \tau $,
$M_{k\text{ }}$ is reversible and $M_k^{-1}(f)(t)=e^{-c_kt}f(t)$. Moreover
\begin{equation*}
M_k^{-1}\circ \tau \circ M_k=\lambda _k\tau ,
\end{equation*}
henceforth
\begin{equation*}
M_k^{-1}\circ (\tau -\lambda _k\operatorname{id})\circ M_k=\lambda _k(\tau -\operatorname{id}),
\end{equation*}
and consequently
\begin{equation*}
M_k^{-1}\circ (\tau -\lambda _k\operatorname{id})^{m_k}\circ M_k=\lambda
_k^{m_k}(\tau -\operatorname{id})^{m_k}.
\end{equation*}
Then
\begin{equation*}
\ker (\tau -\lambda _k\operatorname{id})^{m_k}=\ker (\tau -\operatorname{id})^{m_k}\circ
M_k^{-1}=M_k(\ker (\tau -\operatorname{id})^{m_k}).
\end{equation*}
So, to complete the proof, it suffices to show that for all $n\geq 1$,
\begin{equation*}
\ker (\tau -\operatorname{id})^{n}=\{ t\mapsto \sum_{i=0}^{n-1}
p_i(t)t^{i},\text{ with the $p_i$ continuous and $1$-periodic}\} .
\end{equation*}
Denote by $E_n$ the set of polynomial applications from $\mathbb{R}$ to
$\mathbb{C}$ with degree $\leq n$. Let us verify that $\forall n\geq 1$,
$(\tau -\operatorname{id})(E_n)=E_{n-1}$. Indeed, one has $\forall k\leq n$,
$t\mapsto (t+1)^{k}-t^{k}$ is a polynomial of degree $\leq k-1$, then
\begin{equation*}
(\tau -\operatorname{id})(E_n)\subset E_{n-1.}
\end{equation*}
Since all polynomial which is periodic is a constant function. Then
$\dim \ker (\tau -\operatorname{id})\cap E_n=1$. Consequently, by Rank formula one has
\begin{equation*}
\dim (\tau -\operatorname{id})(E_n)=(\dim E_n)-1=\dim E_{n-1}.
\end{equation*}
So, $(\tau -\operatorname{id})(E_n)=E_{n-1}$, it results that
\begin{equation*}
(\tau -\operatorname{id})^{n}(E_i)=\{0\} \quad \forall i\leq n-1.
\end{equation*}
If $p$ is continuous and $1$-periodic, one has for $f\in E$,
\begin{equation*}
(\tau -\operatorname{id})(pf)(t)=p(t+1)f(t+1)-p(t)f(t)=p(t)(f(t+1)-f(t)).
\end{equation*}
Then
\begin{equation*}
(\tau -\operatorname{id})(pf)=p(\tau -\operatorname{id})(f),
\end{equation*}
it follows that
\begin{equation*}
(\tau -\operatorname{id})^{n}(pf)=p(\tau -\operatorname{id})^{n}(f),\quad
\forall i\leq n-1,\;\forall f_i\in E_i,\;(\tau
-\operatorname{id})^{n}(pf_i)=0,
\end{equation*}
from where, we have that
\begin{equation*}
\{t\mapsto \sum_{i=0}^{n-1} p_i(t)t^{i},
\text{ with the $p_i$ continuous and $1$-periodic}\}
\subset \ker (\tau -\operatorname{id})^{n}.
\end{equation*}
Let us prove the other inclusion by recurrence on $n$.
For $n=1$, one has all element of $\ker (\tau -\operatorname{id})$ is continuous and
$1$-periodic. Let $n\geq 2$, assume that the result is hold for $n-1$,
and let us prove that the result remains true for $n$.
Let $f\in \ker (\tau -\operatorname{id})^{n}$, then $(\tau -\operatorname{id})(f)\in \ker (\tau -\operatorname{id})^{n-1}$
and by recurrent hypothesis one has
$(\tau -\operatorname{id})(f)(t)=\sum_{i=0}^{n-2} p_i(t)t^{i}$,
 but the map $t\mapsto t^{i}$ is in
$E_i=(\tau-\operatorname{id})(E_{i+1})$, then there exists $q_{i+1}\in E_{i+1}$ such that
$t^{i}=(\tau -\operatorname{id})(q_{i+1})(t)$, from which one has
$f-\sum_{i=0}^{n-2} p_iq_{i+1}\in \ker (\tau -\operatorname{id})$. Then there exists a
continuous and $1$-periodic function $p$ such that
$f-\sum_{i=0}^{n-2}p_iq_{i+1}=p$, thus
 $f=p+\sum_{i=0}^{n-2} p_iq_{i+1}$ with $q_{i+1}\in E_{n-1}$, consequently
$f$ is in the required form.
\end{proof}

\section{Asymptotic behavior for solutions of the special difference equation}

\begin{proposition} \label{P12}
 Let $f\in C(\mathbb{R},\mathbb{C})$, then the following three
properties are equivalent:
\begin{itemize}
\item[(1)] Equation \eqref{a} admits a solution $x\in C(\mathbb{R},\mathbb{C})$
such that $\lim_{t\to +\infty } x(t)=0$.

\item[(2)] for all $a\in \mathbb{R}$, the series $\sum_{n\geq 0}f(t+n)$
converges uniformly on $[a,+\infty[$.

\item[(3)] The series $\sum_{n\geq 0} f(t+n)$ converges pointwise on
$\mathbb{R}$ and $\lim_{t\to +\infty }
\sum_{n=0}^{+\infty} f(t+n)=0.$\newline
Under these conditions, $x$ is unique and is given by
$x(t)=-\sum_{n=0}^{+\infty } f(t+n)$.
\end{itemize}
\end{proposition}

\begin{proof}
$(1)\Rightarrow (2)$ One has
$\sum_{k=0}^n  f(t+k)=x(t+n+1)-x(t)$ with
$\lim_{t\to +\infty} x(t)=0$,
then the series $\sum_{n\geq 0} f(t+n)$ converges pointwise on
$\mathbb{R}$ and $x(t)=-\sum_{k=0}^{+\infty} f(t+k)$
which ensures the uniqueness of $x$. On the other hand, for $t\in [
a,+\infty [ $ one has $\overset{+\infty }{\underset{k=n+1}{-\sum }}
f(t+k)=x(t+n+1)$. For $\varepsilon >0$, $\exists $ $A>0$, $\forall t\geq A$,
$|x(t)|\leq \varepsilon $. Let $n_{0}\in \mathbb{N}$ be such that $
a+n_{0}+1\geq A$, then
for all $n\geq n_{0}$, for all $t\in [a,+\infty [ $,
$| \sum_{k=n+1}^{+\infty} f(t+k)| \leq \varepsilon $,
which leads to the uniformly
convergence of the series $\sum_{n\geq 0} f(t+n)$ on $[a,+\infty[$.

$(2)\Rightarrow (1)$ One has that for all $a\in \mathbb{R}$, the series
$\sum_{n\geq 0} f(t+n)$ converges uniformly on $[a,+\infty [ $, and
$t\mapsto f(t+n)$ is continuous on $\mathbb{R}$, then by the continuity
theorem, the map
$x:t\mapsto -\sum_{n=0}^{+\infty} f(t+n)$ is continuous
on $\mathbb{R}$, and one has $\lim_{t\to +\infty} x(t)=0$,
\begin{equation*}
x(t+1)-x(t)=-\sum_{n=1}^{+\infty} f(t+n)
+\sum_{n=0}^{+\infty} f(t+n)=f(t).
\end{equation*}

$(2)\Rightarrow (3)$ for all $a\in \mathbb{R}$, the series
$\sum_{n\geq 0} f(t+n)$ converges uniformly on $[a,+\infty [$.
In particular, the series $\sum_{n\geq 0} f(t+n)$ converges pointwise
on $\mathbb{R}$, in the other part, the uniform convergence on
$\mathbb{R}^{+}$ gives that $\sup_{t\in \mathbb{R}^{+}} |f(t+n)|\to 0$ as
$n\to +\infty$. Then for $\varepsilon >0$, there exists
$n_{0}\in \mathbb{N}$, such that for all $n\geq n_{0}$, and all $t\in \mathbb{R}^{+}$,
$|f(t+n)|\leq \varepsilon $ implies   $|f(t)|\leq\varepsilon $
for all $t\geq n_{0}$, hence $\lim_{t\to +\infty} f(t)=0$; on the
other hand the series $\sum_{n\geq 0} f(t+n)$ converges uniformly
on $\mathbb{R}^{+}$, hence by the inversion limit theorem one has
\begin{equation*}
\lim_{t\to +\infty} \sum_{n=0}^{+\infty } f(t+n)
=\sum_{n=0}^{+\infty } \lim_{t\to +\infty } f(t+n)=0.
\end{equation*}

$(3)\Rightarrow (2)$ Let $a\in \mathbb{R}$ and $\varepsilon >0$, since
$\lim_{t\to +\infty} \sum_{n=0}^{+\infty } f(t+n)=0$, there exists $ A>0$,
$| \sum_{k=0}^{+\infty}f(t+k) | \leq \varepsilon$ for all $t\geq A$.
Let $n_{0}\in \mathbb{N}$ such that $a+n_{0}+1\geq A$ and $t\in [ a,+\infty
[ $, for $n\geq n_{0}$ one has
\begin{equation*}
| \sum_{k=n+1}^{+\infty } f(t+k)|
=| \sum_{k=0}^{+\infty} f(t+1+n+k)|
\leq \varepsilon ,
\end{equation*}
because $t+1+n\geq a+1+n_{0}\geq A$, from what the series
$\sum_{n\geq 0} f(t+n)$ converges uniformly on $[a,+\infty [ $.
\end{proof}

\begin{proposition} \label{P22}
Let $f\in C(\mathbb{R},\mathbb{C})$. Then the following three
properties are equivalent:
\begin{itemize}
\item[(1)] Equation \eqref{a} has a solution
$x\in C(\mathbb{R}, \mathbb{C})$ such that $\lim_{t\to -\infty }x(t)=0$.

\item[(2)] for all $a\in \mathbb{R}$, the series $\sum_{n\geq 1} f(t-n)$
converges uniformly on $]-\infty ,a]$.

\item[(3)] The series $\sum_{n\geq 1} f(t-n)$ converges pointwise on
$\mathbb{ R}$ and $\lim_{t\to-\infty}
\sum_{n=1}^{+\infty} f(t-n)=0$.
Under these conditions $x$ is unique and is defined by
 $x(t)=\sum_{n=1}^{+\infty}f(t-n)$.
\end{itemize}
\end{proposition}

\begin{proof}
Putting $G(t)=x(-t)$ and $F(t)=-f(-t-1)$, the equation can be rewritten as
\begin{equation*}
G(t+1)-G(t)=F(t),\quad \forall t\in \mathbb{R}
\end{equation*}
and the proposition  \ref{P12} yields the conclussion.
\end{proof}

\begin{theorem}[Tauberian Theorem of Hardy \cite{SF}]  \label{thm5}
Let $(u_n)_{n\geq 0}$ be a
complex sequence such that $u_n=O(1/n)$ as $n\to +\infty$,
$s_n=\sum_{k=0}^n u_k$,
$\sigma _n=\frac {1}{n}\sum_{k=0}^{n-1} s_k$.
If $(\sigma_n)_{n\geq 0}$ is convergent, then
$(s_n)_{n\geq 0}$ converges to the same limit.
\end{theorem}

\begin{proposition}[Poisson summation Formula] \label{P31}
Let $f\in C(\mathbb{R}, \mathbb{C})$ such that
for all $a\in \mathbb{R}$, the series $\sum_{n\geq 0} f(t+n)$ converges
uniformly on $[a,+\infty [ $ and the
series $\sum_{n\geq 0} f(t-n)$ is also uniformly convergent on
$]-\infty ,a]$. Then:
\begin{itemize}
\item[(1)] The function $p(t)=\sum_{n\in \mathbb{Z}} f(t+n)$ is defined
and continuous $1$-periodic on $\mathbb{R}$, for all $n\in \mathbb{Z}$,
$\int_{\mathbb{R}}f(t)e^{i2n\pi t}dt$ converges and
\begin{equation*}
c_n(p)=\hat{f}(2\pi n)=\int_{\mathbb{R}}f(t)e^{-i2\pi nt}dt.
\end{equation*}

\item[(2)] If moreover $\hat{f}(2\pi n)=O(1/|n|)$ as $|n|\to +\infty$,
then the Fourier series of $p$ converges pointwise on $\mathbb{R}$ and one has.
 The Poisson summation formula is
\begin{equation*}
\sum_{n\in \mathbb{Z}} f(t+n)=
\sum_{n\in \mathbb{Z}}\sum \hat{f}(2\pi n)e^{i2n\pi t}\quad
\forall t\in \mathbb{R}.
\end{equation*}
\end{itemize}
\end{proposition}

\begin{proof}
By uniform convergence of the series $\sum_{n\geq 0} f(t+n)$ and
$\sum_{n\geq 0} f(t-n)$, one has $p$ is well defined and continuous
on $\mathbb{R}$, and it is clear that $p$ is $1$-periodic,
\begin{equation*}
c_k(p)=\int_{0}^{1}p(t)e^{-2ik\pi t}dt=\int_{0}^{1}\underset{n\in \mathbb{Z
}}{\sum }f(n+t)e^{-2ik\pi t}dt.
\end{equation*}
One has $\sum_{n\geq 0} f(t+n)$ converges uniformly on $[0,1]$.
Since $e^{-2ik\pi t}$ is independent from $n$ and $|e^{-2ik\pi t}|=1$, it follows
that $\sum_{n\geq 0} f(t+n)e^{-2ik\pi t}$ is also uniformly convergent
on $[0,1]$. Then
\begin{align*}
\int_{0}^{N}f(t)e^{-2ik\pi t}dt
&=\sum_{n=0}^{N-1} \int_n^{n+1}f(t)e^{-2ik\pi t}dt\\
&=\sum_{n=0}^{N-1} \int_{0}^{1}f(t+n)e^{-2ik\pi t}dt \\
&=\int_{0}^{1}\sum_{n=0}^{N-1} f(t+n)e^{-2ik\pi t}dt\\
&\to \int_{0}^{1} \sum_{n=0}^{+\infty} f(t+n)e^{-2ik\pi t}dt
\end{align*}
as $N\to +\infty$.
If $A>0$, let $N=E(A)$ (where $E$ denotes the greatest integer function) one
has
\begin{equation*}
\Big| \int_{N}^{A}f(t)e^{-2ik\pi t}dt\Big|
\leq \int_{N}^{A}| f(t)| dt
\leq \int_{N}^{N+1}|f(t)| dt
=\int_{0}^{1}|f(t+N)|dt
\end{equation*}
approaches $0$ because $f(t+N)$ converges uniformly to $0$.
Then we  conclude that $\int_{\mathbb{R}^{+}}f(t)e^{-2ik\pi t}dt$
converges and
\begin{equation*}
\int_{\mathbb{R}^{+}}f(t)e^{-2ik\pi t}dt=\int_{0}^{1}
\sum_{n=0}^{+\infty} f(t+n)e^{-2ik\pi t}dt.
\end{equation*}
In the same manner the uniform convergence on $[0,1]$ of the series
$\sum_{n\geq 1} f(t-n)$ implies that
$\int_{\mathbb{R}^{-}}f(t)e^{-2ik\pi t}dt$ converges and
\begin{equation*}
\int_{\mathbb{R}^{-}}f(t)e^{-2ik\pi t}dt
=\int_{0}^{1}\sum_{n=1}^{+\infty } f(t-n)e^{-2ik\pi t}dt.
\end{equation*}
So $\int_{\mathbb{R}}f(t)e^{-2ik\pi t}dt$ converges and
\begin{equation*}
\int_{\mathbb{R}}f(t)e^{-2ik\pi t}dt
=\int_{0}^{1}\sum_{n\in \mathbb{Z}} f(t+n)e^{-2ik\pi t}dt=c_k(p).
\end{equation*}
Putting $u_{0}=c_{0}$ and for $n\geq 1$,
$u_n=c_ne^{i2\pi nt}+c_{-n}e^{-i2\pi nt}$,
$s_n=\sum_{k=0}^n u_k$,
$s_n$ is the Fourier series of $p$, moreover, since $p$ is continuous, then
from Fejer, we know that $\sigma _n$ (The C\'esaro mean of $s_n)$
converges to $p(t)$ and since
$c_n(p)\underset{|n|\to +\infty }{=} O(\frac{1}{| n| })$,
 $u_n=O(\frac{1}{n})$ from
Hardy one has $s_n$ converges to $p(t)$ which yields the Poisson summation
formula : $p(t)=\sum_{n\in \mathbb{Z}} c_ne^{i2n\pi t}$.
\end{proof}

\begin{remark} \rm
 In the literature, one finds often the main conditions as for example:
there exists $\alpha >1$ such that $f(t)=O(1/|t|^{\alpha})$ as
$|t|\to +\infty$, and $\sum_{n\in \mathbb{Z}}|\hat{f}(2\pi n)| <+\infty $.
Here we have  conditions which are weaker namely:
$\sum_{n\geq 0} f(t+n)$ (resp $\sum_{n\geq 0} f(t-n)$)
converges uniformly on $[a,+\infty [ $ (resp. $]-\infty ,a])$ and $
\hat{f}(2\pi n)=O(1/n)$.
\end{remark}

\begin{proposition} \label{p8}
Let $f\in C(\mathbb{R},\mathbb{C})$, then the following
properties are equivalent:
\begin{itemize}
\item[(1)] Equation \eqref{a} has a solution $x\in C(\mathbb{R},
\mathbb{C})$ such that $\lim_{| t| \to+\infty } x(t)=0$;

\item[(2)] for all $a\in \mathbb{R}$, the series $\sum_{n\geq 0} f(t+n)$
(resp. $\sum_{n\geq 1} f(t-n)$) converges uniformly on $
[a,+\infty [$ (resp. $]-\infty ,a]$) and for all $t\in \mathbb{R}$,
$\sum_{n\in \mathbb{Z}} f(t+n)=0$;

\item[(3)] for all $a\in \mathbb{R}$, the series $\sum_{n\geq 0} f(t+n)$
(resp. $\sum_{n\geq 1} f(t-n)$) converges uniformly on
$[a,+\infty [$ (resp. on $]-\infty ,a]$) and for all $n\in \mathbb{Z}$,
$\int_{\mathbb{R}}f(t)e^{-2in\pi t}dt=0$.
\end{itemize}
Under these conditions $x$ is unique and is given by
\begin{equation*}
x(t)=\sum_{n=1}^{+\infty} f(t-n)=-\sum_{n=0}^{+\infty } f(t+n).
\end{equation*}
\end{proposition}

The above proposition is an immediate consequence of the propositions
\ref{P12}, \ref{P22}  and \ref{P31}.

\begin{proposition} \label{P51}
Let $f\in C(\mathbb{R},\mathbb{C})$ such that $f$ is $C^{1}$ piecewise.

(1) If $f'$ is integrable on $\mathbb{R}^{+}$ and $\int_{\mathbb{R}^{+}}f$ $(t)dt$
converges, then for all $a\in \mathbb{R}$, the
series $\sum_{n\geq 0} f(t+n)$ converges uniformly on $[a,+\infty[$.

(2) If $f'$ is integrable on $\mathbb{R}^{-}$ and $\int_{\mathbb{R}^{-}}f(t)dt$
 converges, then for all $a\in \mathbb{R}$, the
series $\sum_{n\geq 0} f(t-n)$ converges uniformly on $]-\infty,a]$.
\end{proposition}

\begin{proof}
(1) Let $s\in \mathbb{R}$. One has
\begin{align*}
\int_n^{n+1}f(s+t)dt
&=[(t-n-1)f(s+t)]_n^{n+1}-\int_n^{n+1}(t-n-1)f'(s+t)dt \\
&=f(s+n)-\int_n^{n+1}(t-n-1)f'(s+t)dt,
\end{align*}
then
\begin{equation*}
| f(s+n)-\int_n^{n+1}f(s+t)dt| \leq
\int_n^{n+1}|f'(s+t)|dt.
\end{equation*}
Since $f'$ is integrable on $\mathbb{R}^{+}$ and $\int_{\mathbb{R}
^{+}}f(t)dt$ converges, then the series
\begin{equation*}
\sum_{n\geq 0} \int_n^{n+1}|f'(s+t)|dt
\end{equation*}
and
\begin{equation*}
\sum_{n\geq 0} \int_n^{n+1}f(s+t)dt
\end{equation*}
are convergent. Hence the series $\sum_{n\geq 0} f(s+n)$ is also
convergent and
\begin{equation*}
\Big| \overset{+\infty }{\underset{k=n+1}{\sum }}f(s+k)-\int_{n+1+s}^{+
\infty }f(t)dt\big| \leq \int_{n+1+s}^{+\infty }|f'(t)|dt.
\end{equation*}
For $a\in \mathbb{R}$ and $s\in [ a,+\infty [ $, we have
\begin{equation*}
\big| \sum_{k=n+1}^{+\infty } f(s+k)\big|
\leq \big| \int_{n+1+s}^{+\infty }f(t)dt| +\int_{n+1+a}^{+\infty
}\big|f'(t)|dt.
\end{equation*}
Since $\int_{\mathbb{R}^{+}}f(t)dt$ converges, for
$\varepsilon >0$ there exists $A>0$ such that
\[
| \int_{s}^{+\infty }f(t)dt| \leq \varepsilon\quad  \forall s\geq A,.
\]
Let $n_{0}\in \mathbb{N}$ such that $n_{0}+1+a\geq A$, then
for all $n\geq n_{0}$ and all $s\in [ a,+\infty [$,
\[
|\int_{n+1+s}^{+\infty }f(t)dt| \leq \varepsilon .
\]
It results that the series $\sum_{n\geq 0} f(s+n)$ converges
uniformly on $[a,+\infty [$.

(2) follows  from (1) by putting $\varphi (t)=f(-t)$.
\end{proof}

\begin{proposition}\label{P61}
Let $f\in C(\mathbb{R},\mathbb{C})$ such that $f$ is $C^{1}$
piecemeal, $f'$ is integrable on $\mathbb{R}$ and
$\int_{\mathbb{R}}f(t)dt$ converges. Then
$p(t)=\sum_{n\in \mathbb{Z}} f(n+t)$ is
well defined continuous and $1$-periodic on $\mathbb{R}$, for all
$n\in \mathbb{Z}$, $\int_{\mathbb{R}}f(t)e^{i2n\pi t}dt$ converges and
\begin{equation*}
c_n(p)=\hat{f}(2\pi n)=\int_{\mathbb{R}}f(t)e^{-i2\pi nt}dt.
\end{equation*}
The Fourier series of $p$ converges pointwise on $\mathbb{R}$ and one has the
Poisson summation formula:
\begin{equation*}
\sum_{n\in \mathbb{Z}} f(n+t)=
\sum_{n\in \mathbb{Z}} \hat{f}(2\pi n)e^{i2n\pi t}\quad \forall t\in \mathbb{R}.
\end{equation*}
\end{proposition}

\begin{proof}
From Propositions \ref{P31} and \ref{P51}, it suffices to verify that
$c_n(p)=O(1/|n|)$ as $|n|\to +\infty$.
 Since $ f'$ is integrable on $\mathbb{R}$ and $\int_{\mathbb{R}}f(t)dt$
converges, it follows that
$\lim_{|t|\to +\infty } f(t)=0$. An integration by parts
leads to
\begin{equation*}
|c_n(p)|=| \frac{1}{2ni\pi }
\int_{\mathbb{R}} f'(t)e^{-i2n\pi t}dt|
\leq \frac{1}{2| n| \pi }
\int_{\mathbb{R}}| f'(t)| dt,
\end{equation*}
from which, we have that $c_n(p)=O(1/|n|)$ as $|n|\to +\infty$.
\end{proof}

\subsubsection{Characterization of bounded solutions}

Let $\mathcal{B}=\mathbf{C}_{B}(\mathbb{R},\mathbb{C})$ be the space of
bounded and continuous functions from $\mathbb{R}$ to $\mathbb{C}$. We have
seen for $f\in \mathbf{C}(\mathbb{R},\mathbb{C})$, that equation \eqref{a}
has a solution in $\mathcal{B}$ if and only if there exists $c>0$ such that
for all $t\in \mathbb{R}$, and all$ n\in \mathbb{N}$,
$| \sum_{k=0}^n f(t+k)| \leq c$, see \cite{al}.
In the sequel , we consider the case where the period of $f$ is
in $\mathbb{R}\setminus \mathbb{Q}$. Let $C_{T}$ be the space of continuous
and $T$-periodic functions from $\mathbb{R}$ to $\mathbb{C}$.

\begin{proposition}\label{m4p}
Let $f\in C_{T}$ with $T\notin \mathbb{Q}$, then equation \eqref{a}
 has a solution in $\mathcal{B}$ if and only if the sequence $
\sum_{k=0}^n f(k)$ is bounded.
\end{proposition}

\begin{proof}
If \eqref{a} admits a solution in $\mathcal{B}$, then  there exists $c>0$ such
that for all $t\in \mathbb{R}$ and all $n\in \mathbb{N}$,
$| \sum_{k=0}^n f(t+k)| \leq c$ in particular for $t=0$
one has the sequence $\sum_{k=0}^n f(k)$ is bounded.
Reciprocally suppose that there exists $c>0$ such that for all
$n\in \mathbb{N}$, $| \sum_{k=0}^n f(k)| \leq c$,
then
\begin{equation*}
\big| \sum_{k=0}^n f(k+m)\big|
=\big| \sum_{k=m}^{n+m} f(k)\big|
=\big| \sum_{k=0}^{n+m} f(k)-
\sum_{k=0}^{m-1} f(k)| \leq 2c\quad \forall m\in \mathbb{N},
\end{equation*}
and since $f\in C_{T}$, we have
\begin{equation*}
| \sum_{k=0}^n f(k+m-Tq)| \leq 2c\quad \forall q\in \mathbb{N},
\end{equation*}
but $T\notin \mathbb{Q}$ then $\mathbb{N}-T\mathbb{N}$ is dense in
 $\mathbb{R }$, from which
\begin{equation*}
\big|\sum_{k=0}^n f(k+t)\big| \leq 2c,\quad
\forall t\in \mathbb{R},\; \forall n\in \mathbb{N}.
\end{equation*}
It follows that equation \eqref{a} has a solution in $\mathcal{B}$.
\end{proof}

\subsubsection{Existence of periodic solutions}

\begin{lemma} \label{L1}
 Let $g$ be a continuous function which has a period $m\in \mathbb{N}^{\ast }$.
Then
\begin{equation*}
\lim_{n\to +\infty } \frac{1}{n}
\sum_{k=0}^{n-1} g(t+k)=\frac{1}{m}
\sum_{k=0}^{m-1} g(t+k)\quad  \forall t\in \mathbb{R}.
\end{equation*}
\end{lemma}

\begin{proof}
Since $u\mapsto g(u+t)$ has a same period as $g$, it suffices to consider
the case $t=0$. Let $n\in \mathbb{N}^{\ast }$, making the Euclidean division
of $n$ by $m:n=q_nm+r_n$, $0\leq r_n\leq m-1$.
\begin{equation*}
\sum_{k=0}^{n-1} g(k)=\sum_{k=0}^{q_n-1}
\sum_{i=0}^{m-1} g(km+i)
+\sum_{i=0}^{r_n-1} g(q_nm+i)=q_n \sum_{i=0}^{m-1} g(i)
+\sum_{i=0}^{r_n-1} g(i),
\end{equation*}
then
\begin{equation*}
\frac{1}{n}\sum_{k=0}^{n-1} g(k)
=\frac{q_n}{n}\sum_{i=0}^{m-1} g(i)
+\frac{1}{n}\sum_{i=0}^{r_n-1} g(i)\underset{n\to +\infty }{\to }
\frac{1}{m}\sum_{i=0}^{m-1} g(i).
\end{equation*}
\end{proof}

\begin{proposition} \label{P2}
Let $f$ be a continuous function on $\mathbb{R}$, which has a
period $m\in \mathbb{N}^{\ast }$. If  equation \eqref{a} has a continuous
solution $x$ defined on $\mathbb{R}$ and bounded on $\mathbb{R}^{+}$, then
\begin{equation*}
\sum_{k=0}^{m-1} f(t+k)=0\quad \forall t\in \mathbb{R}.
\end{equation*}
\end{proposition}

\begin{proof}
Assume that \eqref{a} has a solution $x$ which is bounded on $\mathbb{R}^{+}$.
Then
\begin{equation*}
\sum_{k=0}^{n-1} f(t+k)=\sum_{k=0}^{n-1} x(t+k+1)-x(t+k)=x(t+n)-x(t),
\end{equation*}
then $\lim_{n\to +\infty} \frac{1}{n}
\sum_{k=0}^{n-1} f(t+k)=0$ and from Lemma \ref{L1} one
has
\begin{equation*}
\sum_{k=0}^{m-1} f(t+k)=0\quad \forall t\in \mathbb{R}.
\end{equation*}
\end{proof}

\begin{corollary} \label{coro14}
Let $f$ be a continuous function defined on $\mathbb{R}$, which is not zero
with period $1$. Then \eqref{a} has not a continuous solution
$x$ on $\mathbb{R}$ and  bounded on $\mathbb{R}^{+}$.
\end{corollary}

\begin{proposition}\label{P3}
Let $f$ be a continuous function on $\mathbb{R}$, with a period
$m\in \mathbb{N}^{\ast}$ and
$m\geq 2$. Then  \eqref{a} has a solution $x$ which is
continuous on $\mathbb{R}$ and bounded on $\mathbb{R}^{+}$ if and only if $f$
is of the form
\begin{equation*}
f(t)=\sum_{k=1}^{m-1} f_k(t)\exp (\frac{2ik\pi }{m}t),
\end{equation*}
where the $f_k$ are continuous and $1$-periodic. In this case all
solutions of \eqref{a} are $m$-periodic of the form
\begin{equation*}
f(t)=f_{0}(t)+\sum_{k=1}^{m-1} f_k(t)\frac{\exp (
\frac{2ik\pi }{m}t)}{\exp(\frac{2ik\pi }{m})-1}
\end{equation*}
where $f_{0}$ is continuous and $1$-periodic.
\end{proposition}

\begin{proof}
If $f(t)=\sum_{k=1}^{m-1} f_k(t)\exp (\frac{2ik\pi }{m}t)$, then it is clear
 that
\begin{equation*}
f_{0}(t)=\sum_{k=1}^{m-1} f_k(t)
\frac{\exp (\frac{2ik\pi }{m}t)}{\exp(\frac{2ik\pi }{m})-1}
\end{equation*}
is a particular solution of equation \eqref{a}, since $\ker (\tau -\operatorname{id})$ is
formed by continuous and $1$-periodic functions, then $x$ is of the request
form and is $m$-periodic.
 Conversely, if equation \eqref{a} has a
bounded solution defined on $\mathbb{R}^{+}$, then from  proposition
 \ref{P2},  for all $t\in \mathbb{R}$, $\sum_{k=0}^{m-1} f(t+k)=0$.
On the other hand,
\begin{equation*}
1+X+X^{2}+..+X^{m-1}=\prod_{k=1}^{m-1} (X-e^{\frac{2ik\pi }{m}})
\end{equation*}
and from  proposition  \ref{P1},
\begin{equation*}
f(t)=\sum_{k=1}^{m-1} f_k(t)\exp (\frac{2ik\pi }{m}t)
\end{equation*}
where the $f_k$ are any continuous and $1$-periodic.
\end{proof}

\begin{proposition} \label{prop16}
Let $f:\mathbb{R\to R}$ be a continuous function which is $T$-periodic with
$T=\frac{n}{p}\in \mathbb{Q}^{+\ast }$, $n\wedge p=1$. Then the following
statements are equivalent:
\begin{itemize}
\item[(1)] Equation \eqref{a} has a solution $x$ continuous on $\mathbb{R}$
and bounded on $\mathbb{R}^{+}$;

\item[(2)] for all $t\in \mathbb{R}$, $\sum_{k=0}^{p-1} f(t+k)=0$;

\item[(3)] for all $k\in \mathbb{Z}$, $c_{2k\pi }(f)=0$.
\end{itemize}
Under these conditions equation \eqref{a} has a unique $T$-periodic
solution $x$ such that for all $k\in \mathbb{Z}$, $c_{2k\pi }(x)=0$.
Moreover $x$ is given by
\begin{equation*}
x(t)=\frac{1}{p} \sum_{k=1}^{p} kf(t+k-1).
\end{equation*}
\end{proposition}

\begin{proof}
 Let $x_{1},x_{2}$ be two solutions of equation \eqref{a} satisfying
$c_{2k\pi }(x_{1})=c_{2k\pi }(x_{2})=0$ for all $k\in \mathbb{Z}$.
Then $x=x_{1}-x_{2}$ is $1$-periodic and for all $k\in \mathbb{Z}$,
$c_{2k\pi }(x)=0$. Then $x=0$ and  $x_{1}=x_{2}$.

$(1)\Longrightarrow (2)$ If $f$ is $T=\frac{n}{p}$ periodic, then it is $p$ periodic
and the end of the proof results from proposition \ref{P2}.

$(2)\Leftrightarrow (3)$ The application $x:t\mapsto \sum_{k=0}^{p-1}
f(t+k)$ is $1$ periodic, then $x=0\Leftrightarrow
 \forall m\in \mathbb{Z}$, $c_{2m\pi }(x)=0$. However
\begin{equation*}
c_{2m\pi }(x)=\overset{p-1}{\underset{k=0}{\sum }}e^{2mk\pi i}c_{2m\pi
}(f)=pc_{2m\pi }(f)
\end{equation*}
then $(2)\Leftrightarrow (3)$.

$(2)\Longrightarrow (1)$ One has
\begin{align*}
\sum_{k=0}^{p-1} (X^{k}-1)
&=(X-1) \sum_{k=1}^{p-1}  \sum_{j=0}^{k-1} X^{j}
=(X-1) \sum_{j=0}^{p-2} \sum_{k=j+1}^ {p-1} X^{j}\\
&= (X-1)\sum_{j=0}^{p-2} (p-1-j)X^{j}
=(X-1) \sum_{k=1}^{p-1}  (p-k)X^{k-1} \\
&=(X-1) \sum_{k=1}^{p} (p-k)X^{k-1},
\end{align*}
it follows that
\begin{equation*}
\sum_{k=0}^{p-1} \tau ^{k}=p\operatorname{id}+(\tau -\operatorname{id})
\sum_{k=1}^{p} (p-k)\tau ^{k-1}.
\end{equation*}
Then
\begin{equation*}
0=\overset{p-1}{\underset{k=0}{\sum }}f(t+k)=pf(t)+(\tau -\operatorname{id})\overset{p}{
\underset{k=1}{\sum }}(p-k)f(t+k-1).
\end{equation*}
But
\begin{equation*}
\sum_{k=1}^{p} pf(t+k-1)=p\sum_{k=0}^{p-1} f(t+k)=0,
\end{equation*}
hence
\begin{equation*}
f(t)=\frac{1}{p}(\tau -\operatorname{id})\sum_{k=1}^{p} kf(t+k-1).
\end{equation*}
Consequently
\begin{equation*}
x(t)=\frac{1}{p}\sum_{k=1}^{p} kf(t+k-1),
\end{equation*}
is a solution of  \eqref{a} which is continuous on $\mathbb{R}$ and
$T$-periodic, in particular it is bounded on $\mathbb{R}^{+}$. Moreover one
has that for all $m\in \mathbb{Z}$,
\begin{equation*}
c_{2m\pi }(x)=\frac{1}{p} \sum_{k=1}^{p} ke^{2m(k-1)i\pi}c_{2m\pi }(f)=0,
\end{equation*}
because $c_{2m\pi }(f)=0$.
\end{proof}

\begin{proposition} \label{prop17}
Suppose that $f$ is continuous $T$-periodic with mean value zero for $
T\notin \mathbb{Q}$. Then  $\eqref{a}$\ has a $T$-periodic
solution if and only if $\frac{1}{n}\sum_{k=0}^{n-1} (n-k)\tau ^{k}f$
converges uniformly on $\mathbb{R}$. Under these conditions,
\begin{equation*}
x_{0}(t)=-\lim_{n\to +\infty}  \frac{1}{n} \sum_{k=0}^{n-1}(n-k)f(t+k),
\end{equation*}
is a unique $T$-periodic solution of \eqref{a} with mean value zero.
\end{proposition}

\begin{proof}
One has $f$ is $T$ periodic with $T\notin \mathbb{Q}$ then for all
$\lambda \in 2\pi \mathbb{Z}^{\ast }$, $c_{\lambda }(f)=0$.
In fact if there exists $k\in \mathbb{Z}$ such that
$\lambda =\frac{2k\pi }{T}$, we will have $T\in \mathbb{Q}$ which is absurd.
Moreover one has $f$ is mean value zero then
for all $\lambda \in 2\pi \mathbb{Z}$, $c_{\lambda }(f)=0$ namely
$f\in F$. From the almost periodic case
\cite{alalmost} equation \eqref{a} has a solution in $AP(\mathbb{R},\mathbb{C})$
if and only if the sequence $\frac{1}{N+1}\overset{N}{\underset{n=0}{\sum }}s_n$
converges uniformly on $\mathbb{R}$, where
 $s_n= \sum_{k=0}^n f(t+k)$ and under these conditions
$x_{0}(t)=-\lim_{N\to +\infty} \frac{1}{N+1}\overset{N}{\underset{n=0}{\sum }}s_n(t)$
is the unique solution in $F$ of equation \eqref{a} and as $f$ is
$T$-periodic, then $x$ is also, on the other hand the set of functions which are
$T$-periodic with mean value zero is include in $F$, (since $T\notin
\mathbb{Q}$) then $x$ is the unique $T$-periodic solution with mean value
zero. As
\begin{equation*}
\frac{1}{N+1} \sum_{n=0}^N s_n(t)=\frac{1}{N+1}
\sum_{k=0}^N (N+1-k)f(t+k),
\end{equation*}
then $x_{0}$ is written as
\begin{equation*}
x_{0}(t)=-\lim_{n\to +\infty}  \frac{1}{n} \sum_{k=0}^{n-1} (n-k)f(t+k).
\end{equation*}
\end{proof}

\begin{remark} \label{rmk18} \rm
The set of $T$-periodic solutions of equation \eqref{a} is of the form $
x_{0}+c$ where $c\in \mathbb{C}$.
\end{remark}

\subsubsection{Characterization of sequences which are Fourier coefficients
of periodic and continuous functions}

Let $f\in C_{T}$, $\omega _n=\frac{2n\pi }{T}$,
 $(c_n(f))_{n\in \mathbb{Z}}$ the family of Fourier coefficients of $f$:
\begin{equation*}
c_n(f)=\frac{1}{T}\int_{0}^{T}f(t)e^{-i\omega _nt}dt.
\end{equation*}
From F\'ejer results, we know that the C\'esaro mean of the
sequence
$S_n(f)(t)=\sum_{k=-n}^n c_k(f)e^{i\omega _kt}$
converges uniformly to $f$ on $\mathbb{R}$.
Now, we give the proof of the reciprocal result.

\begin{proposition} \label{mp1}
 Let $(\lambda _n)_{n\in \mathbb{Z}}$ be a family of complex
numbers such that the C\'esaro mean of the sequence
$\underset{k=-n}{\overset{n}{\sum }}\lambda _ke^{i\omega _kt}$ is uniformly
convergent on $\mathbb{R}$, then its limit $x$ is the unique function in
$C_{T}$, such that $\forall n\in \mathbb{Z}$, $c_n(x)=\lambda _n$.
\end{proposition}

\begin{proof}
Denote $S_n(t)=\sum_{k=-n}^n \lambda _ke^{i\omega _kt}$ and
$x(t)=\lim_{N\to +\infty } \frac{1}{N}\sum_{n=0}^{N-1} S_n(t)$.
Since $\frac{1}{N}\sum_{n=0}^{N-1} S_n\in C_{T}$ and the uniform
convergence of the series, then $f\in C_{T}$. Now let us verify that
for all $p\in \mathbb{Z}$,
$c_{p}(x)=\lambda _{p}$. In fact, for $n\geq |p|$, $c_{p}(S_n)=\lambda_{p}$, then
\begin{equation*}
c_{p}(\frac{1}{N}\sum_{n=| p| }^{N-1} S_n)=\frac{N-| p| }{N}\lambda _{p}\quad
\forall N>|p|,
\end{equation*}
and as $\frac{1}{N}\sum_{n=| p| }^{N-1} S_n$ converges uniformly to $f$, then
\begin{equation*}
\lim_{N\to +\infty } c_{p}(\frac{1}{N}\sum_{n=| p| }^{N-1} S_n)=c_{p}(x)
\end{equation*}
from what, for all $p\in \mathbb{Z}$, $c_{p}(x)=\lambda _{p}$. The
uniqueness of $x$ results from that the elements of $C_{T}$ which have the
same Fourier coefficients are equal .
\end{proof}

Let $T\in \mathbb{R\setminus Q}$ which is non negative, $f\in C_{T}$: We
will give a characterization of the solutions $x\in C_{T}$ of $\eqref{a}$ by
the Fourier coefficients of $f$.

\begin{proposition}\label{mp0}
Equation \eqref{a} has a solution $f\in C_{T}$ if and only
if $c_{0}(f)=0$ and the C\'esaro mean of the sequence
$\sum_{1\leq |k|\leq n} \frac{c_k(f)}{e^{i\omega _k}-1}e^{i\omega _kt}$ is
uniformly convergent, its limit $x_{0}$ is the unique solution of \eqref{a}
satisfying $c_{0}(x_{0})=0$. The solutions in $C_{T}$ of \eqref{a} are of
the form $x=c+x_{0}$, where $c$ is a constant.
\end{proposition}

\begin{proof}
Note that for $k$ $\neq 0$, $\frac{c_k(f)}{e^{i\omega _k}-1}$ is well
definite because $T\notin \mathbb{Q}$.

$(\Rightarrow )$ Let $x\in C_{T}$ be a solution of \eqref{a}, then
\begin{equation*}
c_{0}(x)=c_{0}(t\to x(t+1))-c_{0}(x)=c_{0}(x)-c_{0}(x)=0,
\end{equation*}
for $k\neq 0$ one has $e^{i\omega _k}c_k(x)-c_k(x)=c_k(f)$, from
where $c_k(x)=\frac{c_k(f)}{e^{i\omega _k}-1}$ and then
\begin{equation*}
\sum_{1\leq |k|\leq n} \frac{c_k(f)}{e^{i\omega _k}-1}e^{i\omega _kt}
=\sum_{1\leq |k|\leq n} c_k(x)e^{i\omega _kt}.
\end{equation*}
From Fejer, we conclude that the C\'esaro mean of the following sequence
\newline
$\sum_{1\leq |k|\leq n} \frac{c_k(f)}{e^{i\omega _k}-1}e^{i\omega _kt}$
converges uniformly to $x_{0}=x-c_{0}(x)$. If moreover
$c_{0}(x)=0$, then the Fourier coefficients of $x$ are unique; which gives
the uniqueness of the solution $x\in C_{T}$ such that $c_{0}(x)=0$, and the
solutions in $C_{T}$ of $\eqref{a}$ are of the form $x=c+x_{0}$ where $c$ is
a constant.

$(\Leftarrow )$ We have the C\'esaro mean of the  sequence
\[
S_n= \sum_{1\leq |k|\leq n} \frac{c_k(f)}{e^{i\omega _k}-1}
e^{i\omega _kt}
\]
 is uniformly convergent, then thinks to  proposition
\ref{mp1} its limit $x_{0}\in C_{T}$ is such that
$c_{0}(x_{0})=0$ and for all $k\in \mathbb{Z}^{\ast }$,
$c_k(x_{0})=\frac{c_k(f)}{e^{i\omega _k}-1}$, furthermore one has $c_{0}(f)=0$,
then the application $t\mapsto x_{0}(t+1)-x_{0}(t)$ has the same Fourier coefficients
as $f$, from which $x_{0}(t+1)-x_{0}(t)=f(t)$.
\end{proof}

\subsubsection{Characterization of irrational periodic solutions}

In this section, we  discuss the characterization of irrational periodic
solution. Let $T$ be a nonnegative irrational number, then $\mathbb{R}/T
\mathbb{Z}$ is a metric space with respect to the distance defined by
\begin{equation*}
d(\bar{x},\bar{y})=|\exp (\frac{2\pi i}{T}x)-\exp (\frac{2\pi i}{T}y)|.
\end{equation*}
We denote
\begin{equation*}
\mathbb{U}=\{z\in \mathbb{C}:\text{ }|z|=1\}.
\end{equation*}

\begin{proposition} \label{PC1}
(1) The map $\phi :\mathbb{R\to R}/T\mathbb{Z}$ defined
by $\phi (x) =\bar{x}$ is continuous on $\mathbb{R}$.

(2) The map $r:\mathbb{R\to }[0,T[$, $x\mapsto x-T[ \frac{x}{T}]$
 ($[\cdot] $ denotes the greatest integer function) is
$T$-periodic and $\overline{r(x)}=\overline{x}$.

(3) The metric space $(\mathbb{R}/T\mathbb{Z}$, $d)$ is compact.

(4) The map $h:\mathbb{R}/T\mathbb{Z\to U}$, $\bar{x}\mapsto \exp (
\frac{2\pi i}{T}x)$ is an homeomorphism.

(5) Let $\bar{x}_{0}\in \mathbb{R}/T\mathbb{Z}$, then $\{\bar{x}_{0}+\bar{n}$,
$n\in \mathbb{N}\}$ is dense in $\mathbb{R}/T\mathbb{Z}$.
\end{proposition}

\begin{proof}
(1) results from $x\mapsto \bar{x}$ begin $\frac{2\pi }{T}$ Lipschitzian, thus
continuous.

(2) is a consequence of the facts that
$E(\frac{x}{T})\leq \frac{x}{T}<1+E(\frac{x}{T})$ and $x\mapsto E(x)$ is
$1$-periodic.

(3) Let $(\bar{x}_n)_n$ be a sequence of $\mathbb{R}/T\mathbb{Z}$,
$r(x_n)$ is bounded, which has a convergent subsequence
$r(x_{\varphi(n)})$, let $\ell $ be its limit, and by continuity of
$x\mapsto \bar{x}$
one has $\bar{x}_{\varphi (n)}$ tends to $\bar{\ell}$.

(4) It is clear that $h$ is bijective and one has
$d(\bar{x}_n,\bar{x})=|h(\bar{x}_n)-h(\bar{x})|$, then $h$ is bicontinuous.

(5) As $T$ is a nonnegative irrational number, then $\mathbb{N}-T\mathbb{N}$
is dense in $\mathbb{R}$, it follows that $x_{0}+\mathbb{N}-T\mathbb{N}$ is
also dense, and by continuity of $x\mapsto \bar{x}$, we have that
$\{\bar{x}_{0}+\bar{n}$, $n\in \mathbb{N}\}$ is dense in $\mathbb{R}/T\mathbb{Z}$.
\end{proof}

\noindent\textbf{Notation} We denote by $\mathcal{P}_{T}$ the space of continuous
functions defined from $\mathbb{R}$ to $\mathbb{C}$ which are $T$-periodic.

\begin{proposition} \label{PC2}
The map $\psi :\mathcal{P}_{T}\to C(\mathbb{R}/T\mathbb{Z}$, $\mathbb{C})$,
 defined by $f\longmapsto \tilde{f}$, where $\tilde{f}(\bar{x})=f(x)$,
is a bijection.
\end{proposition}

\begin{proof}
$\tilde{f}$ is well defined since $f$ is $T$-periodic. Let us prove that it
is continuous: In deed if $\bar{x}_n$ tends to $\bar{x}$. Then using (4)
of proposition \ref{PC1} one has that for all $k\in \mathbb{Z}$,
$[\exp (\frac{2\pi i}{T}x_n)]^{k}$ tends to
$[\exp (\frac{2\pi i}{T} x)]^{k}$, hence for all trigonometric polynomial
$P\in \mathcal{P}_{T}$, one has $P(x_n)$ goes to $P(x)$ and since $f$
is a uniform limit of trigonometric polynomials, then $f(x_n)$ tends to $f(x)$,
from where $\tilde{f}$ is continuous, it is clear that $\psi $ is one to one.
It remains to prove that it is surjective, let
$g\in C(\mathbb{R}/T\mathbb{Z}, \mathbb{C})$, putting $f=g\circ p$ where
$p:x\mapsto \bar{x}$, then
\begin{equation*}
f(x+T)=g(\overline{x+T})=g(\bar{x})=f(x)
\end{equation*}
and $f$ is continuous as composition of continuous functions, then
$g=\tilde{ f}$, so $\psi $ is surjective.
\end{proof}

\begin{corollary} \label{mcor}
 Let $u\in \mathcal{P}_{T}$. Then $f$ is a solution in $\mathcal{P}_{T}$ of
the equation
$f(x+1)-f(x)=u(x)$ if and only if $\tilde{f}$ is a solution in
 $C(\mathbb{R} /T\mathbb{Z}$,$\mathbb{C})$ of the following equation:
\begin{equation}
g(\bar{x}+\bar{1})-g(\bar{x})=\tilde{u}(\bar{x}).  \label{***}
\end{equation}
\end{corollary}

\begin{proposition} \label{PC3}
Let $u\in \mathcal{P}_{T}$ and $g$ a solution in
$C(\mathbb{R}/T \mathbb{Z}, \mathbb{C})$ of  equation \eqref{***}.
Let $K=\{(\bar{x},f(\bar{x}))$, for $\bar{x}\in \mathbb{R}/T\mathbb{Z\}}$.
Then $K$ is a non empty compact subset which is stable by the map
\begin{equation*}
s:(\bar{x},y)\mapsto (\bar{x}+\bar{1},y+u(x))
\end{equation*}
and $K$ is minimal for the inclusion, namely for all non empty compact
$K'$ which is invariant under $s$, one has $K'\subset K\Rightarrow K'=K$.
\end{proposition}

\begin{proof}
One has $K$ is a range of the compact $\mathbb{R}/T\mathbb{Z}$ by a
continuous map $\bar{x}\mapsto (\bar{x},g(\bar{x}))$, then $K$ is a non
empty compact set. Let $(\bar{x},g(\bar{x}))\in K$. Then
\begin{equation*}
s(\bar{x},g(\bar{x}))=(\bar{x}+\bar{1},g(\bar{x})+u(x))
=(\bar{x}+\bar{1},g( \bar{x}+\bar{1})),
\end{equation*}
from where $K$ is invariant under $s$. Let $K'$ a non empty compact
stable by $s$ such that
$K'\subset K$, $(\bar{x}_{0},f(\bar{x}_{0}))\in K'$, hence
for all $n\geq 1$, $s^{n}((\bar{x}_{0},f(\bar{x}_{0})))\in K'$,
namely
$(\bar{x}_{0}+\bar{n},f(\bar{x}_{0}+\bar{n}))\in K'$.
We have $ \left\{ \bar{x}_{0}+\bar{n},\text{ }n\in \mathbb{N}\right\} $ is dense in
$\mathbb{R}/T\mathbb{Z}$, but as $g$ is
continuous and $K'$ is closed, then for all $\bar{x}\in \mathbb{R}/T\mathbb{Z}$,
$(\bar{x},f(\bar{x}))\in K'$, namely $K\subset K'$ so $K'=K$.
\end{proof}

\begin{proposition} \label{mpr1}
Let $u\in \mathcal{P}_{T}$ and $K$ a non empty compact
invariant under the application
$s:(\bar{x},y)\mapsto (\bar{x}+\bar{1},y+u(x))$ and $K$ is minimal for
the inclusion. Then $\eqref{***}$ has a
solution $f$ in $C(\mathbb{R}/T\mathbb{Z}$,$\mathbb{C})$ such that
 $K=\{(\bar{x},f(\bar{x}))$, for all $\bar{x}\in \mathbb{R}/T\mathbb{Z\}}$.
\end{proposition}

\begin{proof}
One has $K\neq \emptyset $, let $(\bar{x}_{0},y_{0})\in K$, stable by $s$,
then for all $n\geq 1$, $s^{n}(\bar{x}_{0},y_{0})\in K$, namely
\begin{equation*}
(\bar{x}_{0}+\bar{n},y_{0}+\sum_{k=0}^{n-1} u(x+k))\in K,
\end{equation*}
one has $p:(\bar{x},y)\mapsto \bar{x}$ is continuous and
$\bar{x}_{0}+\bar{n} \in p(K)$ which is compact, but
$\{\bar{x}_{0}+\bar{n}$, $n\in \mathbb{N}\}$
is dense in $\mathbb{R}/T\mathbb{Z}$. Then
 $\mathbb{R}/T\mathbb{Z\subset }p(K)$ from where
$p(K)=\mathbb{R}/T\mathbb{Z}$. Let $(\bar{x},y)\in K$ and
$d\in \mathbb{C}^{\ast }$. Consider
$f:\mathbb{R}/T\mathbb{Z\times C\to R}/T\mathbb{Z\times C}$ defined by
\[
f(\bar{w},\lambda )=(\bar{w},\lambda +d).
\]
One has $f$ is continuous and commutes with $s$. Then $f(K)\cap K$ is also a
compact stable by $s$. As $K$ is minimal then
\begin{equation*}
f(K)\cap K=K\quad \text{or}\quad f(K)\cap K=\emptyset ,
\end{equation*}
similarly $f^{-1}(K)\cap K$ is a compact stable by $s$, leads to
\begin{equation*}
f^{-1}(K)\cap K=K\quad \text{or}\quad f^{-1}(K)\cap K=\emptyset
\end{equation*}
which is equivalent to $K\cap f(K)$ $=\emptyset $.
Then if we assume that $f(K)\cap K\neq \emptyset $, we will have at once
\begin{equation*}
f(K)\cap K=K\quad \text{and}\quad f^{-1}(K)\cap K=K
\end{equation*}
which gives
\begin{equation*}
K\subset f(K)\quad \text{and}\quad K\subset f^{-1}(K),
\end{equation*}
then $f(K)=K$. It follows that for all $n\geq 1$, $f^{n}(K)=K$ and for all
$n\geq 1$, $(\bar{x},y+nd)\in K$ which contradicts the boundedness of $K$,
then $K\cap f(K)=\emptyset $, hence for all $d\in \mathbb{C}^{\ast }$,
$(\bar{x},y+d)\notin K$. So $(\bar{x},y)\in K$ and
$(\bar{x},z)\in K$ implies that $y=z$.
Moreover $p(K)=\mathbb{R}/T\mathbb{Z}$, then for all $\bar{x}\in
\mathbb{R}/T\mathbb{Z}$, there exists a unique $y\in \mathbb{C}$ such that
$(\bar{x},y)\in K$; so we will define a map $g:\mathbb{R}/T\mathbb{Z\to C}$
such that
\begin{equation*}
K=\{(\bar{x},g(\bar{x})):\text{ such that }\bar{x}\in \mathbb{R}/T\mathbb{Z\}}.
\end{equation*}
Let us prove that $g$ is continuous, indeed if $\bar{x}_n$ goes to
$\bar{a}$, as $K$ is compact then its projections are compact, then
$g(\bar{x}_n)$ is a sequence in the compact. To prove that it converges to
$g(\bar{a})$ it suffices to prove that $g(\bar{a})$ is the alone adhesion
 value of $g(\bar{x}_n)$. Let $\ell $ an adhesion value of $g(\bar{x}_n)$,
$\ell =\lim_{n\to +\infty } g(\bar{x}_{\varphi (n)})$, one has
\begin{equation*}
(\bar{x}_{\varphi (n)},g(\bar{x}_{\varphi (n)}))
\underset{n\to +\infty }{\to }(\bar{a},\ell )\in K,
\end{equation*}
hence $\ell =g(\bar{a})$. As $K$ is invariant under $s$, then
$(\bar{x}+\bar{1},g(\bar{x})+u(x))\in K;$ from where
\begin{equation*}
g(\bar{x}+\bar{1})=g(\bar{x})+u(x).
\end{equation*}
\end{proof}

\begin{proposition} \label{PC4}
 Let $T\in \mathbb{R}^{+\ast }\backslash \mathbb{Q}$, $u\in
\mathcal{P}_{T}$, such that the sequence $\sum_{k=0}^n u(k)$ is bounded,
then for all $(x_{0},y_{0})\in \mathbb{R}\times \mathbb{C}$,
 the equation $f(x+1)-f(x)=u(x)$ has a unique solution in
$\mathcal{P}_{T} $ such that $f(x_{0})=y_{0}$.
\end{proposition}

\begin{proof}
Uniqueness: Let $f$ and $g$ be two solutions, then
$f-g\in \mathcal{P}_{T}\cap \mathcal{P}_{1}$, but the set of periods of
 $f-g$ is a group $G$, and as it contains $T\mathbb{Z+Z}$ which is dense
in $\mathbb{R}$ and that $G $ is closed then $G=\mathbb{R}$, hence $f-g$
is constant, but $(f-g)(x_{0})=0 $ consequently $f-g=0$ which leads
to the uniqueness.

Existence: Let $s:(\bar{x},y)\mapsto (\bar{x}+\bar{1},y+u(x))$ and
$A=\{s^n(\overline{0},0), n\in \mathbb{N}\}$, one has that
for all $n\geq 1$,
$s^{n}(\overline{0},0)=(\bar{n}, \sum_{k=0}^{n-1} u(k))$,
then there exists $c>0$ such that
$A\subset (\mathbb{R}/T \mathbb{Z})\times \{z\in \mathbb{C}, |z|\leq c\}$
which is compact and non empty then $L=\bar{A}$ is a non empty compact and
as $A$ is invariant under $s$ and $s$ is continuous, then $L$ is compact
and invariant under $s$. Let
\begin{equation*}
E=\{K\text{ non empty compact subset of }(\mathbb{R}/T\mathbb{Z)\times C}
\text{ stable by }s\},
\end{equation*}
one has $E$ is non empty because $L\in E$. Let $(K_i)_{i\in I}$ a family
totally ordinate of $E$, then $\cap_{i\in I} K_i$ is also
compact invariant under $s$, if it was empty, it could exist a finite subset
$J$ of $I$, such that $\cap_{i\in J} K_i=\emptyset $ and as the
family $(K_i)_{i\in I}$ is a family totally ordinate of $E$, then
there exists $j\in J$ such that $\cap_{i\in J} K_i=K_{j}$;
 hence $K_{j}=\emptyset $ absurd. Then
$\cap_{i\in I} K_i\neq \emptyset $ and
by the Zorn lemma one has $E$ has a minimal element then thanks to
proposition \ref{mpr1}, the equation $g(\bar{x}+\bar{1})-g(
\bar{x})=\tilde{u}(\bar{x})$ has a solution in
$C(\mathbb{R}/T\mathbb{Z},\mathbb{C})$ and from the corollary
\ref{mcor}  the equation $f(x+1)-f(x)=u(x)$ has a solution
 $f_{0}$ in $\mathcal{P}_{T}$. Then if we
put $f(x)=f_{0}(x)+y_{0}-f_{0}(x_{0})$ then $f$ is an answer to the question.
\end{proof}

\section{Examples}

In this section, we give some examples to demonstrate the results obtained
in previous sections.

\begin{example}[Solutions which go to $0$] \label{examp27} \rm
Let $\lambda \in \mathbb{R}^{\ast}$, $I$ an interval in $\mathbb{R}$ and
$\varphi :I\mapsto \mathbb{C}$ such that
$u\mapsto \frac{\varphi (u)}{e^{i\lambda u}-1}$ is integrable on $I$.
Then for all $n\in \mathbb{Z}$,
\begin{equation*}
\int_{\mathbb{R}}\Big(\int_{I}\varphi
(u)e^{it(\lambda u+2n\pi )}du\Big)dt=0
\quad\text{and}\quad
\sum_{n\in \mathbb{Z}} \int_{I}\varphi (u)e^{i\lambda (t+n)u}du=0.
\end{equation*}
In fact, we consider the equation \eqref{a}, where
\begin{equation*}
f(t)=\int_{I}\varphi (u)e^{i\lambda tu}du.
\end{equation*}
Since $u\mapsto \frac{\varphi (u)}{e^{i\lambda u}-1}$ is integrable on $I$,
then $\varphi $ is also. Thus $f$ is defined and continuous on $\mathbb{R}$,
on the other hand, thanks to Riemann Lebesgue lemma, we have
$\lim_{|t|\to +\infty } f(t)=0$, and
\begin{equation*}
x(t)=\int_{I}\frac{\varphi (u)}{e^{i\lambda u}-1}e^{i\lambda tu}du
\end{equation*}
is the unique solution of equation \eqref{a} such that
$\lim_{|t|\to +\infty } x(t)=0$.
Hence, from proposition  \ref{p8}, it follows that for all $n\in \mathbb{Z}$,
\begin{equation*}
\int_{\mathbb{R}}f(t)e^{2in\pi t}dt=0\quad \text{and}\quad
\sum_{n\in \mathbb{Z}} f(t+n)=0.
\end{equation*}
\end{example}

\begin{example}[Examples of periodic solutions] \label{examp28} \rm
(1) Thanks to proposition \eqref{mp1}, if the C\'esaro mean of
\begin{equation*}
s_n(t)=\sum_{1\leq |k|\leq n} \frac{c_k(f)}{e^{i\omega _k}-1}
e^{i\omega _kt}
\end{equation*}
converges uniformly, in particular if $s_n$ is uniformly convergent, then
its limit $x_{0}$ is a solution of equation \eqref{a}.

(2) Let $(\alpha _n)_{n\in \mathbb{N}^{\ast }}$ be a nonnegative
non increasing sequence such that $\alpha _n=o(1/n)$ (for example
$\alpha _n=\frac{1}{n\ln (n+1)}$),
\begin{equation*}
f(t)=\sum_{n=1}^{+\infty} \alpha _n\sin (n)\cos (2t+1)n,
\end{equation*}
then equation \eqref{a} has as a $\pi$-periodic solution
\begin{equation*}
x_{0}(t)=\sum_{n=1}^{+\infty} \frac{\alpha _n\sin (2nt)}{2}
\end{equation*}
which has mean value zero. In fact we have
\begin{equation*}
\alpha _n\sin (n)\cos (2t+1)n=\frac{1}{2}\alpha _n\sin (n)(e^{(2t+1)in}+e^{-(2t+1)in})
\end{equation*}
Hence, if we assume in first time that the series which defines $f$ is
uniformly convergent, we will have $c_{0}(f)=0$, and for $n\geq 1$,
\begin{equation*}
c_n(f)=\frac{1}{2}\alpha _n\sin (n)e^{in},\text{ }c_{-n}(f)=\overline{c_n(f)}.
\end{equation*}
So
\begin{equation*}
c_n(x_{0})=\frac{c_n(f)}{e^{2in}-1}=\frac{\alpha _n\sin (n)}{2}\frac{
e^{in}}{e^{2in}-1}=-\frac{i}{4}\alpha _n
\end{equation*}
Thus we obtain
\begin{align*}
c_n(x_{0})e^{2int}+c_{-n}(x_{0})e^{-2int}
&=2\operatorname{Re} (c_n(x_{0})e^{2int})\\
&=2\frac{\alpha _n}{4}\operatorname{Re}(-ie^{2int})
=\frac{\alpha _n\sin (2nt)}{2}
\end{align*}
it follows that
\begin{equation*}
x_{0}(t)=\sum_{n=1}^{+\infty} \frac{\alpha _n\sin (2nt)}{2}
\end{equation*}
and we have
\begin{align*}
x_{0}(t+1)-x_{0}(t)
&=\sum_{n=1}^{+\infty} \alpha _n \frac{\sin (2nt+2n)-\sin (2nt)}{2} \\
&=\sum_{n=1}^{+\infty} \frac{2\alpha _n\sin (\frac{2nt+2n-2nt}{2})
\cos (\frac{2nt+2n+2nt}{2})}{2} \\
&=\sum_{n=1}^{+\infty }  \alpha _n\sin (n)\cos (2t+1)n=f(t).
\end{align*}
To complete the proof, it suffices to verify the uniform convergence on
 $\mathbb{R }$ of the series
\begin{equation*}
\sum_{n\geq 1} \alpha _n\sin (2nt).
\end{equation*}
For this, it suffices to prove the uniform convergence of
$\sum_{n\geq 1} \alpha _n\sin (nt)$ on $[0,\pi ]$.
For $m>n\geq 1$; and $t\in [0;\pi ]$ one has
\begin{align*}
&\sum_{k=n}^n 2\alpha _k\sin (kt)\sin (\frac{t}{2}) \\
&=\sum_{k=n}^m \alpha _k(\cos (k-\frac{1}{2})t-\cos (k+
\frac{1}{2})t) \\
&=\sum_{k=n-1}^{m-1} \alpha _{k+1}\cos (k+\frac{1}{2})t-
\sum_{k=n}^m \alpha _k\cos (k+\frac{1}{2})t \\
&=\alpha _n\cos (n-\frac{1}{2})t-\alpha _{m}\cos (m+\frac{1}{2})t
+\sum_{k=n}^{m-1} (\alpha _{k+1}-\alpha _k)\cos (k+\frac{1}{2})t.
\end{align*}
Hence
\begin{equation*}
\big| \overset{m}{\underset{k=n}{\sum }}2\alpha _k\sin (kt)\sin \frac{t}{2}\big|
\leq \alpha _n+\alpha _{m}+\sum_{k=n}^{m-1} (\alpha _k-\alpha _{k+1})
\end{equation*}
which implies
\begin{equation*}
\big| \overset{m}{\underset{k=n}{\sum }}\alpha _k\sin (kt)\big|
\sin \frac{t}{2}\leq \alpha _n
\end{equation*}
which allows us to see that the series
\begin{equation*}
\sum_{n\geq 1} \alpha _n\sin (nt)
\end{equation*}
is pointwise convergent on $[0,\pi ]$; and if $m$ goes to $+\infty $
and $t\in ]0,\pi ]$; one has
\begin{equation*}
\big| \sum_{k=n}^{+\infty } \alpha _k\sin (kt)|
\leq \frac{\alpha _n}{\sin \frac{t}{2}}
\end{equation*}
Let $t\in ]0,\pi ]$, and put
$E=[ \frac{\pi }{t}] $, the greatest integer part of
$\frac{\pi }{t}$, and $u_n=\sup_{k\geq n} k\alpha _k$. Then we
\begin{gather*}
\begin{aligned}
\big| \sum_{k=n+E}^{+\infty} \alpha _k\sin (kt)\big|
&\leq \frac{\alpha _{n+E}}{\sin \frac{t}{2}}
\leq \frac{\alpha _{n+E}}{\frac{2}{\pi }\frac{t}{2}}\\
&\leq \frac{\pi }{t}\alpha _{n+E}
\leq (1+E)\alpha _{n+E} \\
&\leq (n+E)\alpha _{n+E}\leq u_n
\end{aligned} \\
\end{gather*}
which leads to
\begin{equation*}
| \sum_{k=n}^{+\infty } \alpha _k\sin (kt)| \leq (1+\pi )u_n
\end{equation*}
and the inequality remains valid for $t=0$. Tt follows that the series $
\sum_{n\geq 1} \alpha _n\sin (nt)$ converges uniformly on $[0,\pi ]$.
\end{example}

\subsection*{Acknowledgements}
 The authors express their sincere thanks to the
anonymous referee for their careful reading,  helpful comments, and valuable
suggestions which improve this article.

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\end{document}