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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 13, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/13\hfil Remark on periodic BVPs]
{Remark on periodic boundary-value problem for second-order linear ordinary
differential equations}

\author[M. Dosoudilov\'a, A. Lomtatidze \hfil EJDE-2018/13\hfilneg]
{Monika Dosoudilov\'a, Alexander Lomtatidze}

\address{Monika Dosoudilov\'a \newline
Institute of Automation and Computer Science,
Faculty of Mechanical Engineering,
Brno University of Technology,
Technick\'a 2, 616 69 Brno, Czech Republic}
\email{dosoudilova@fme.vutbr.cz}

\address{Alexander Lomtatidze \newline
Institute of Mathematics,
Faculty of Mechanical Engineering,
Brno University of Technology,
Technick\'a 2, 616 69 Brno, Czech Republic\newline
Institute of Mathematics,
Czech Academy of Sciences,branch in Brno,
\v{Z}i\v{z}kova 22,  616 62 Brno, Czech Republic}
\email{lomtatidze@fme.vutbr.cz}

\dedicatory{Communicated by Pavel Drabek}

\thanks{Submitted October 3, 2017. Published January 10, 2018.}
\subjclass[2010]{34B05, 34C25}
\keywords{Second-order linear equation; periodic boundary value problem; 
\hfill\break\indent unique solvability}

\begin{abstract}
 We establish conditions for the unique solvability of periodic
 boundary value problem for second-order linear equations.
 We make more precise a result proved in \cite{a862}.
\end{abstract}

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

\section{Introduction}
\label{sec:introduction}

Consider the periodic boundary-value problem
\begin{equation} \label{eq:1}
u''=p(t)u+q(t);\quad u(0)=u(\omega),\ u'(0)=u'(\omega),
\end{equation}
where $p,q: [0,\omega]\to\mathbb{R}$ are Lebesgue integrable
functions. By a solution of given in \eqref{eq:1} equation, 
as usual, we understand a function $u\in\mathit{AC}^1([0,\omega])$ such that
for almost all $t\in[0,\omega]$. 

\begin{definition} \label{def:1} \rm
We say that the function $p\in L([0,\omega])$ belongs to the set $V^-(\omega)$
(resp. $V^+(\omega)$) if for every $u\in\mathit{AC}^1([0,\omega])$ satisfying
$$
u''(t)\geq p(t)u(t)\quad\text{for a.e. } t\in[0,\omega],\quad u(0)=u(\omega),\quad
u'(0)=u'(\omega),
$$
the inequality
\begin{equation}
\label{eq:2}
u(t)\leq0\quad\text{for } t\in[0,\omega]\quad(\text{resp.}\ u(t)\geq0\quad\text{for }
t\in[0,\omega])
\end{equation}
is fulfilled.
\end{definition}

It is clear that if $p\in V^-(\omega)$ (resp. $p\in V^+(\omega)$), then the
homogeneous problem
$$
u''=p(t)u;\quad u(0)=u(\omega),\ u'(0)=u'(\omega)
$$
has no nontrivial solution. Consequently, by virtue of Fredholm's
alternative, the problem \eqref{eq:1} is uniquely solvable. Moreover,
if $q(t)\geq0$ for $t\in[0,\omega]$, then the unique solution $u$ of
the problem \eqref{eq:1} satisfies \eqref{eq:2}.

It is also evident that if $p\in V^-(\omega)$ (resp. $p\in V^+(\omega)$) and the
functions $u,v\in\mathit{AC}^1([0,\omega])$ satisfy differential inequalities
$$
u''(t)\geq p(t)u(t),\quad v''(t)\leq p(t)v(t)\quad\text{for a.e. } t\in[0,\omega]
$$
and boundary conditions
$$
u^{(i)}(0)-u^{(i)}(\omega)=v^{(i)}(0)-v^{(i)}(\omega),\quad
i=0,1,
$$
then the inequality
$$
u(t)\leq v(t)\quad\text{for } t\in[0,\omega]\quad(\text{resp.}\ u(t)\geq v(t)\quad\text{for }
t\in[0,\omega])
$$
holds.

Properties of the sets $V^-(\omega)$ and $V^+(\omega)$ plays a crucial
role in the theory of periodic boundary value problems for nonlinear
equations (see, e.\,g., \cite{a862,b}). Therefore, it is desirable to
establish sufficient conditions for the inclusion $p\in V^-(\omega)$,
resp. $p\in V^+(\omega)$.  One can find several integral conditions in
\cite{a862}. 

\begin{theorem}[{\cite[Theorem~11.1]{a862}}] \label{thm:1}
Let $p\not\equiv0$ and
\begin{equation} \label{eq:3}
\|[p]_-\|_1\leq\frac{\|[p]_+\|_1}{1+\frac{\omega}{4}\,\|[p]_+\|_1}\,.
\end{equation}
Then $p\in V^-(\omega)$.
\end{theorem}


The main result of this article makes more precise
Theorem~\ref{thm:1}. In particular, it covers also the case when
$\|[p]_-\|_1\geq 4/\omega $.

Below we use the following notation:
$\mathbb{R}=\,]-\infty,+\infty[\,$. For $x\in\mathbb{R}$, we put
$[x]_+=\frac{1}{2}(|x|+x)$ and $[x]_-=\frac{1}{2}(|x|-x)$.

Let $\omega>0$ and $\lambda\in\,]0,\frac{1}{2}]$. Then
$$
\Delta_{\omega}(\lambda)
:=\big[\frac{1-2\lambda}{2\omega(1-\lambda)}
\big]^{\frac{1-\lambda}{\lambda}}.
$$

The set $\mathit{AC}^1([a,b])$ consists of absolutely continuous functions
$u: [a,b]\to\mathbb{R}$  whose first derivative is also absolutely continuous
on $[a,b]$.
%%%
\marginpar{I modified this sentence. Please check it}
%%%%
The set $L([a,b])$ consists of Lebesgue integrable functions
$f: [a,b]\to\mathbb{R}$.
If $f\in L([a,b])$ and $\lambda\in\,]0,\frac{1}{2}]$, then we
put
\[
\|f\|_{\lambda}=\Big(\int_a^b|f(s)|^{\lambda}\,\mathrm{d}{s}
\Big)^{1/\lambda}.
\]
By $L_{\omega}$ we denote the set of $\omega$-periodic functions
$f: \mathbb{R}\to\mathbb{R}$ such that $f\in L([0,\omega])$.
Now we are able to formulate main results.

\begin{theorem} \label{thm:2}
Let $p\not\equiv0$, $\lambda\in\,]0,\frac{1}{2}[\,$, and
\begin{gather}\label{eq:4}
\|[p]_-\|_1<\frac{4}{\omega}+
\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda},\\[1mm]
\label{eq:5}
\begin{aligned}
\|[p]_-\|_1
&\leq \|[p]_+\|_1\left(1-\frac{\omega}{4}\,\|[p]_-\|_1+
\frac{\omega}{4}\,\Delta_{\omega}(\lambda)
\|[p]_-\|_{\lambda}\right)\\
&\quad+\frac{\omega}{4}\,\Delta_{\omega}(\lambda)
\|[p]_+\|_{\lambda}\|[p]_-\|_1.
\end{aligned}
\end{gather}
Then the inclusion $p\in V^-(\omega)$ holds.
\end{theorem}

\begin{remark} \label{rem:1} \rm
It is not difficult to verify that if \eqref{eq:3} holds then
\eqref{eq:4} and \eqref{eq:5} are fulfilled. Indeed,
it follows from \eqref{eq:3} that
$\|[p]_-\|_1<4/\omega$. Hence, \eqref{eq:4} holds.
On the other hand, \eqref{eq:3} is equivalent to the inequality
$\|[p]_-\|_1+\frac{\omega}{4}\|[p]_+\|_1\|[p]_-\|_1\leq\|[p]_+\|_1$,
i.\,e., $\|[p]_-\|_1\leq\|[p]_+\|_1(1-\frac{\omega}{4})\|[p]_-\|_1$
and consequently, \eqref{eq:5} holds. Thus,
Theorem~\ref{thm:2} generalizes Theorem~\ref{thm:1}. On the other
hand, since $\Delta_{\omega}(1/2)=0$, conditions \eqref{eq:4}
and \eqref{eq:5} with $\lambda=1/2$ are equivalent to
\eqref{eq:3}. In other words, one can regard  Theorem~\ref{thm:1} as
``limit case'' of Theorem~\ref{thm:2}.
\end{remark}

\begin{corollary} \label{cor:1}
Let $p\not\equiv0$ and $\lambda\in\,]0,1/2[\,$. Let, moreover,
one of the following two items be fulfilled:
\begin{itemize}
\item[(i)]
$\|[p]_-\|_1\leq 4/\omega$ and
$\|[p]_+\|_1\|[p]_-\|_{\lambda}+\frac{4}{\omega}\,\|[p]_+\|_{\lambda}\geq
\frac{16}{\omega^2\Delta_{\omega}(\lambda)}\,$;

\item[(ii)]
$\|[[p]_-]\|_1<\frac{4}{\omega}+\Delta_{\omega}(\lambda)
\|[p]_-\|_{\lambda}$ and $\|[p]_+\|_{\lambda}\geq
\frac{4}{\omega\Delta_{\omega}(\lambda)}\,$.
\end{itemize}
Then the inclusion $p\in V^-(\omega)$ holds.
\end{corollary}

To be more concrete, put $\lambda=1/3$. Then
$\Delta_{\omega}(\lambda)=1/(16\omega^2)$ and conditions of
Corollary~\ref{cor:1} reads as follows:
\begin{itemize}
\item[(i)]
$\|[p]_-\|_1\leq 4/\omega$ and
$\|[p]_+\|_1\|[p]_-\|_{1/3}+\frac{4}{\omega}\,
\|[p]_+\|_{1/3} \geq16^2$;

\item[(ii)]
$\|[[p]_-]\|_1<\frac{4}{\omega}+\frac{1}{16\omega^2}
\|[p]_-\|_{1/3}$ and $\|[p]_+\|_{1/3}\geq64\omega$.
\end{itemize}

We postpone the proof of Theorem~\ref{thm:2}  until Section~\ref{sec:proofs},
after some auxiliary propositions stated in
Section~\ref{sec:auxiliary-statements}.

\section{Auxiliary statements}
\label{sec:auxiliary-statements}

First of all for convenience of the reader,  we recall some known
results. 

\begin{definition} \label{def:2} \rm
We say that the function $p\in L_{\omega}$ belongs to the set $D(\omega)$ if the
problem
$$
u''=p(t)u;\quad u(\alpha)=0,\ u(\beta)=0
$$
has no nontrivial solution for any $\alpha<\beta$ satisfying
$\beta-\alpha<\omega$.
\end{definition}

\begin{proposition}[{\cite[Theorem~9.3]{a862}}]
\label{prop:1}
Let $p\in L_{\omega}$,such that $p\not\equiv0$, and
$\int_0^{\omega}p(s)\,\mathrm{d}{s}\leq0$. Then $p\in V^+(\omega)$ if and only if
$p\in D(\omega)$.
\end{proposition}

\begin{proposition}[{\cite[Lemma~2.7]{a862}}]
\label{prop:2}
Let $p\in V^+(\omega)$, $q\in L([0,\omega])$, $q(t)\geq0$ for
$t\in[0,\omega]$, and $q\not\equiv0$. Then the (unique) solution $u$
of the problem \eqref{eq:1} satisfies $u(t)>0$ for $t\in[0,\omega]$.
\end{proposition}

\begin{proposition}[{\cite[Theorem~8.3]{a862}}]
\label{prop:3}
Let $p\in L([0,\omega])$. Then the inclusion $p\in V^-(\omega)$ holds if and
only if there exists a~positive function $\gamma\in\mathit{AC}^1([0,\omega])$
satisfying
$$
\gamma''(t)\leq p(t)\gamma(t)\quad\text{for a.e. } t\in[0,\omega],\quad
\gamma(0)\geq\gamma(\omega),\quad
\frac{\gamma'(\omega)}{\gamma(\omega)}\geq\frac{\gamma'(0)}{\gamma(0)}\,,
$$
and
$$
\gamma(0)-\gamma(\omega)+\frac{\gamma'(\omega)}{\gamma(\omega)}-
\frac{\gamma'(0)}{\gamma(0)}+
\operatorname{meas}\{t\in[0,\omega]:\gamma''(t)<p(t)\gamma(t)\}>0.
$$
\end{proposition}

Let $f\in L([a,a+\omega])$, then we define
\begin{equation}
\label{eq:6}
\begin{aligned}
G_a(f)(t)
&=(a+\omega-t)\int_a^t(s-a)f(s)\,\mathrm{d}{s}\\
&\quad+(t-a)\int_t^{a+\omega}(a+\omega-s)f(s)\,\mathrm{d}{s}\quad\text{for }
t\in[a,a+\omega].
\end{aligned}
\end{equation}

\begin{proposition}
\label{prop:4}
Let $\lambda\in\,]0,\frac{1}{2}[\,$, $f\in L([a,a+\omega])$, and
$f(t)\geq0$ for $t\in[a,a+\omega]$. Then we have the estimates
\begin{gather}
\label{eq:7}
G_a(f)(t)\leq(t-a)(a+\omega-t)\Big(\|f\|_1-\Delta_{\omega}(\lambda)
\|f\|_{\lambda}\Big)\quad\text{for } t\in[a,a+\omega],\\[1mm]
\label{eq:10}
G_a(f)(t)\geq(t-a)(a+\omega-t)\Delta_{\omega}(\lambda)\|f\|_{\lambda}\quad\text{for }
t\in[a,a+\omega]\,.
\end{gather}
\end{proposition}

\begin{proof}
By H\"older's inequality, we have 
\begin{align*}
\int_a^tf^{\lambda}(s)\,\mathrm{d}{s}
&=\int_a^t\big[(s-a)f(s)\big]^{\lambda}
(s-a)^{-\lambda}\,\mathrm{d}{s}\\
&\leq\Big(\frac{1-\lambda}{1-2\lambda}\Big)^{1-\lambda}
(t-a)^{1-2\lambda}
\Big(\int_a^t(s-a)f(s)\,\mathrm{d}{s}\Big)^{\lambda}\quad\text{for } t\in[a,a+\omega].
\end{align*}
Hence,
$$
\int_a^t(s-a)f(s)\,\mathrm{d}{s}\geq
\Big(\frac{1-2\lambda}{1-\lambda}\Big)^{\frac{1-\lambda}{\lambda}}
(t-a)^{-\frac{1-2\lambda}{\lambda}}\Big(\int_a^tf^{\lambda}(s)\,\mathrm{d}{s}
\Big)^{1/\lambda}
$$
for $t\in[a,a+\omega]$. Analogously,
$$
\int_t^{a+\omega}(a+\omega-s)f(s)\,\mathrm{d}{s}\geq
\Big(\frac{1-2\lambda}{1-\lambda}\Big)^{\frac{1-\lambda}{\lambda}}
(a+\omega-t)^{-\frac{1-2\lambda}{\lambda}}\Big(\int_t^{a+\omega}f^{\lambda}(s)
\,\mathrm{d}{s}\Big)^{1/\lambda}.
$$
Consequently,
\begin{equation} \label{eq:8}
\begin{aligned}
G_a(f)(t)&\geq
\Big(\frac{1-2\lambda}{1-\lambda}\Big)^{\frac{1-\lambda}{\lambda}}
(t-a)(a+\omega-t)\Big[
\frac{1}{(t-a)^{\frac{1-\lambda}{\lambda}}}\Big(\int_a^tf^{\lambda}(s)
\,\mathrm{d}{s}\Big)^{1/\lambda}\\
&\quad+\frac{1}{(a+\omega-t)^{\frac{1-\lambda}{\lambda}}}\Big(
\int_t^{a+\omega}f^{\lambda}(s)\,\mathrm{d}{s}\Big)^{1/\lambda}\Big]\\[1mm]
&\geq\Big(\frac{1-2\lambda}{\omega(1-\lambda)}
\Big)^{\frac{1-\lambda}{\lambda}}(t-a)(a+\omega-t)\\
&\quad\times\Big[\Big(\int_a^tf^{\lambda}(s)\,\mathrm{d}{s}\Big)^{1/\lambda}
+\Big(\int_t^{a+\omega}f^{\lambda}(s)\,\mathrm{d}{s}\Big)^{1/\lambda}
\Big]\quad\text{for } t\in\,]a,a+\omega[\,.
\end{aligned}
\end{equation}
On the other hand, it is clear that
\begin{equation}
\label{eq:9}
x^{1/\lambda}+(A-x)^{1/\lambda}\geq
\frac{1}{2^{\frac{1-\lambda}{\lambda}}}\,A^{1/\lambda}\quad\text{for } x\in[0,A].
\end{equation}
Estimate \eqref{eq:10} now follows from \eqref{eq:8} in view of
\eqref{eq:9}.

In the same way one can show that
\begin{equation} \label{eq:11}
H_a(f)(t)\geq(t-a)(a+\omega-t)\Delta_{\omega}(\lambda)\|f\|_{\lambda}\quad\text{for }
t\in[a,a+\omega],
\end{equation}
where
\begin{align*}
H_a(f)(t)&:=(a+\omega-t)\int_a^t(t-s)f(s)\,\mathrm{d}{s}\\
&\quad+(t-a)\int_t^{a+\omega}(s-t)f(s)\,\mathrm{d}{s}\quad\text{for }
t\in[a,a+\omega].
\end{align*}
By direct calculations one can easily verify that
$$
G_a(f)(t)=(t-a)(a+\omega-t)\|f\|_1-H_a(f)(t)\quad\text{for } t\in[a,a+\omega].
$$
Hence, in view of \eqref{eq:11}, we get \eqref{eq:7}.
\end{proof}

\section{Proof of main result}
\label{sec:proofs}

\begin{proof}[Proof of Theorem~\ref{thm:2}]
Extend the function $p$ periodically and denote it by the same
letter. Suppose that $[p]_-\not\equiv0$ since otherwise it is known
(see, e.\,g., Theorem~\ref{thm:1}) that $p\in V^-(\omega)$. In view of
\eqref{eq:4} and \cite[Theorem~1.2]{a882}, we have that
$-[p]_-\in D(\omega)$. Hence, by virtue of Proposition~\ref{prop:1}, the
inclusion $-[p]_-\in V^+(\omega)$ holds as well. Denote by $\gamma$ a solution
of the problem
\begin{equation} \label{eq:12}
\gamma''=-[p(t)]_-\gamma+[p(t)]_+\,;\quad
\gamma(0)=\gamma(\omega),\quad \gamma'(0)=\gamma'(\omega).
\end{equation}
In view of \eqref{eq:5}, it is clear that $[p]_+\not\equiv0$ and
consequently, by  Proposition~\ref{prop:2}, we have
$$
\gamma(t)>0\quad\text{for } t\in[0,\omega].
$$
It is also evident that $\gamma\not\equiv\mathit{Const}$. Now we
show that
\begin{equation} \label{eq:13}
\gamma(t)>1\quad\text{for } t\in[0,\omega].
\end{equation}
Put
$$
m:=\min\big\{\gamma(t):t\in[0,\omega]\big\},\quad
M:=\max\big\{\gamma(t):t\in[0,\omega]\big\}.
$$
Extend the function $\gamma$ periodically and denote it by the same
letter. Then there exists $a\in[0,\omega[\,$ such that
$$
\gamma(a)=m,\quad \gamma(a+\omega)=m.
$$
It is cleat that the function $\gamma$ is a solution of the Dirichlet
problem
\begin{equation}
\label{eq:14}
\gamma''=-[p(t)]_-\gamma+[p(t)]_+\,;\quad
\gamma(a)=m,\quad \gamma(a+\omega)=m\,.
\end{equation}
 By direct calculations one can easily verify that
\begin{equation}
\label{eq:15}
\gamma(t)=m+\frac{1}{\omega}\,G_a([p]_-\gamma)(t)-\frac{1}{\omega}\,
G_a([p]_+)(t)\quad\text{for } t\in[a,a+\omega],
\end{equation}
where $G_a$ is defined by \eqref{eq:6}. By 
Proposition~\ref{prop:4}, we obtain
\begin{align*}
G_a([p]_-\gamma)(t)&\leq MG_a([p]_-)(t)\\
&\leq M(t-a)(a+\omega-t)\Big(
\|[p]_-\|_1-\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)\quad\text{for }
t\in[a,a+\omega]
\end{align*}
and
$$
G_a([p]_+)(t)\geq\Delta_{\omega}(\lambda)(t-a)(a+\omega-t)
\|[p]_+\|_{\lambda}\quad\text{for } t\in[a,a+\omega].
$$
Hence, from \eqref{eq:15} it follows that
\begin{align*}
\gamma(t)\leq m+\frac{(t-a)(a+\omega-t)}{\omega}\Big(
M\Big(\|[p]_-\|_1-\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)
-\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}\Big)
\end{align*}
for $t\in[a,a+\omega]$. Taking now into account that
$\gamma\not\equiv\mathit{Const.}$ we get from the latter inequality
that
$$
M\Big(\|[p]_-\|_1-\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)
-\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}>0
$$
and consequently
\begin{equation}
\label{eq:16}
\gamma(t)<m+\frac{\omega}{4}\big(
M\Big(\|[p]_-\|_1-\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)
-\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}\big)
\end{equation}
for $t\in[a,a+\omega]\setminus\{t_0\}$, where
$t_0= a+\frac{\omega}{2}$.

In view of \eqref{eq:5}, it is clear that
\begin{align*}
&m+\frac{\omega}{4}\Big(
M\Big(\|[p]_-\|_1-\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)
-\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}\Big)\\
&=m-1+1-\frac{\omega}{4}\,\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}+
\frac{\omega}{4}\,M\Big(\|[p]_-\|_1-
\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)\\[1mm]
&\leq m-1+\frac{\omega}{4}\,M\Big(\|[p]_-\|_1-
\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)\\
&\quad+\frac{\|[p]_+\|_1}{\|[p]_-\|_1}\Big(1-
\frac{\omega}{4}\,\|[p]_-\|_1+\frac{\omega}{4}\,\Delta_{\omega}(\lambda)
\|[p]_-\|_{\lambda}\Big).
\end{align*}
From \eqref{eq:16} it follows that
\begin{equation}
\label{eq:17}
\gamma(t)<m-1+\frac{\|[p]_+\|_1}{\|[p]_-\|_1}+\frac{\omega}{4}\Big(
M-\frac{\|[p]_+\|_1}{\|[p]_-\|_1}\Big)\Big(\|[p]_-\|_1-
\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)
\end{equation}
for $t\in[a,a+\omega]\setminus\{t_0\}$. On the other hand,
\eqref{eq:16} implies 
\begin{equation} \label{eq:18}
m\geq M\Big(1-\frac{\omega}{4}\Big(\|[p]_-\|_1-
\Delta_{\omega}(\lambda)\|[p]_-\|_{\lambda}\Big)\Big)+
\frac{\omega}{4}\,\Delta_{\omega}(\lambda)\|[p]_+\|_{\lambda}\,.
\end{equation}
From \eqref{eq:12} it follows that
\begin{equation}
\label{eq:19}
\int_0^{\omega}[p(s)]_+\,\mathrm{d}{s}=\int_0^{\omega}[p(s)]_-\gamma(s)\,\mathrm{d}{s}
\end{equation}
and consequently
$$
M\geq\frac{\|[p]_+\|_1}{\|[p]_-\|_1}\,.
$$

If $M>\frac{\|[p]_+\|_1}{\|[p]_-\|_1}$ then, in view of \eqref{eq:4}
and \eqref{eq:5}, inequality \eqref{eq:18} implies that $m>1$ and
consequently, \eqref{eq:13} holds. Let now
$M=\frac{\|[p]_+\|_1}{\|[p]_-\|_1}$. Then, in view of \eqref{eq:17},
we have
$$
\gamma(t)<m-1+\frac{\|[p]_+\|_1}{\|[p]_-\|_1}\quad\text{for }
t\in[a,a+\omega]\setminus\{t_0\},
$$
which, together with \eqref{eq:19} and the condition $[p]_-\not\equiv0$,
imply
$$
\|[p]_+\|_1<(m-1)\|[p]_-\|_1+\|[p]_+\|_1.
$$
Hence, $m>1$. Thus, we have proved that \eqref{eq:13} holds.

Now it follows from \eqref{eq:12}, in view of \eqref{eq:13}, that the
function $\gamma$ satisfies conditions of Propositions~\ref{prop:3}
and therefore, $p\in V^-(\omega)$.
\end{proof}

\subsection*{Acknowledgments}
M. Dosoudilov\'a was supported by grant FSI-S-4785.
A. Lomtatidze was supported by 
NETME CENTER PLUS (L01202) and by RVO: 67985840.

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\bibitem{b}
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\bibitem{a862}
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\end{thebibliography}

\end{document}