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\begin{document}
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\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 2 (2008), no. 2, 76--84\\
$\frac{\rule{4.55in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\
$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]}


\title[Hardy inequality with one negative parameter]{The Hardy inequality with one negative parameter}

\author[A. Kufner, K. Kuliev,
Kulieva]{A. Kufner$^1$, K. Kuliev$^{2}$$^*$ and G. Kulieva$^3$}

\address{$^{1}$ Mathematical Institute, Academy of Sciences of the  Czech
Republic, \v{Z}itna 25, 11567 Praha 1, Czech Republic}

\email{\textcolor[rgb]{0.00,0.00,0.84}{kufner@math.cas.cz}}

\address{$^{2}$ Department of Mathematics, University of West Bohemia,
Univerzitn\'{\i} 22, 30614 Pilsen, Czech Republic }
\email{\textcolor[rgb]{0.00,0.00,0.84}{komil@kma.zcu.cz}}


\address{$^{3}$ Department of Mathematics, University of West Bohemia,
Univerzitn\'{\i} 22, 30614 Pilsen, Czech Republic}
\email{\textcolor[rgb]{0.00,0.00,0.84}{kulievag@mail.ru}}

\dedicatory{Dedicated to Professor Josip E. Pe\v{c}ari\'{c}\\
\vspace{.5cm}{\rm Submitted by C. Park}}

\subjclass[2000]{Primary 26D10, 26D15; Secondary 47B38.}

\keywords{Inequalities, Hardy-type inequalities, weights, negative
powers.}

\date{Received: 19 February 2008; Accepted: 3 June 2008.\newline \indent $^{*}$ Corresponding author}

\begin{abstract}
In this paper, necessary and sufficient conditions for the validity
of the Hardy inequality for the case $q<0,\, p>0 $ and for the case
$q>0,\, p<0$ are derived.
\end{abstract} \maketitle


\section{Introduction and preliminaries}
\noindent The classical Hardy inequality
\begin{equation}\label{In.1}
\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}\leq C\left(\int_{a}^{b}
f(x)^{p}dx\right)^{\frac{1}{p}}
\end{equation}
for all $f\geq0,$ where $u,\, v$ are weight functions, is almost
completely described for
$p,q$  such that%
$$
p\geq1, \, q>0
$$
%
(see \cite{Kuf.Malig.Per.}, \cite{Kuf.Per.}, \cite{Opic.Kuf.}),
while for $p,\, q$ such that
%
$$
0<p<1, \, q>1
$$
%
it is known that inequality (\ref{In.1}) doesn't hold (see \cite{Kuf.Per.}, p.46).\\


\noindent The so called \textit{reverse Hardy inequality}
\begin{equation}\label{Re.In.1}
\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}\geq C\left(\int_{a}^{b}
f(x)^{p}dx\right)^{\frac{1}{p}}
\end{equation}
was studied in \cite{Bees.Heinig.}  for
%
$$
0<p,q<1\quad \mbox{and}\quad p,q<0;
$$
%
the second case was described  in \cite{Prokhorov} and the case for
$$
-\infty<q\leq p<0
$$
was described in \cite{Kuf.Kul.}.\\


\noindent Here, we want to consider parameters $p,q$ which satisfy
either
%
$$
p<0,\, q>0
$$
%
or
%
$$
p>0,\, q<0.
$$
%
It will be shown that in the first case, the reverse inequality
(\ref{Re.In.1}) hold (see Theorem \ref{the-m1}) while in the second
case, the reverse inequality (\ref{Re.In.1}) holds for $0<p< 1,$ $q<0$ (see Theorem
\ref{rev.the-m2}) and the Hardy inequality (\ref{In.1}) holds  for
$p\geq 1, q<0$ (see Theorem  \ref{the-m2}). The results can be
extended to the "adjoint" inequalities
\begin{equation}\label{Re.In.13}
\left(\int_{a}^{b}\left(\int_{x}^{b} f(t) v(t)dt\right)^q
u(x)dx\right)^{\frac{1}{q}}\leq C\left(\int_{a}^{b}
f(x)^{p}dx\right)^{\frac{1}{p}}
\end{equation}
and
\begin{equation}\label{Re.In.12}
\left(\int_{a}^{b}\left(\int_{x}^{b} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}\geq C\left(\int_{a}^{b}
f(x)^{p}dx\right)^{\frac{1}{p}}
\end{equation}
(see Remark \ref{remark }).\\
 \noindent
 The negative powers $p, q$ force us to work with
functions having values from the interval $[0,\,+\infty].$
Therefore, we define the following arithmetics:
\begin{equation*}
\begin{cases}
0+(+\infty)=a+(+\infty)=a\cdot(+\infty)=\frac{a}{0}=+\infty, & a\in (0,\,+\infty];\\
0\cdot(+\infty)=\frac{a}{+\infty}=0, & a\in [0,\,+\infty);\\
0^{-\alpha}=(+\infty)^{\alpha}=+\infty, \quad 0^{\alpha}=(+\infty)^{-\alpha}=0, & \alpha\in (0,\, +\infty).
\end{cases}
\end{equation*}

\section{Main results}\label{sec1}

\noindent Let us denote
\begin{equation*}\label{eq:1.2}
A(t):=\left(\int_a^tv^{p'}(x)dx\right)^{\frac{1}{p'}}\left(\int_t^bu(x)dx\right)^{\frac{1}{q}},
\quad p'=\frac{p}{p-1}.
\end{equation*}
Then we can formulate the following theorems:
\begin{theorem}\label{the-m1}
Let $p<0 $ and $ q>0$. Then inequality (\ref{Re.In.1}) holds if and
only if there exists $\tau\in (a,b)$ such that
\begin{equation}\label{eq:1.3}
A(\tau)>0.
\end{equation}
Moreover,
\begin{itemize}
\item[i)]\, if
$$
0< A^*:=\sup\limits_{(a,b)} A(t)<\infty,
$$
and $C$ is the best possible constant of inequality (\ref{Re.In.1})
then $A^*\leq C;$
\item[ii)]\, if
$$
A^*=\infty,
$$
then the best constant of inequality (\ref{Re.In.1}) does not exist,
more precisely, the left hand side of (\ref{Re.In.1}) is infinite
for all positive functions for which
$\left(\int_a^bf^p(t)dt\right)^{\frac{1}{p}}>0.$
\end{itemize}
\end{theorem}
\begin{proof}
Let $\tau\in (a,b)$ be arbitrary. Then
\begin{equation*}
\begin{split}
J:=&\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q} u(x)dx\geq
\int_{\tau}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q} u(x)dx\\
\geq& \int_{\tau}^{b}\left(\int_{a}^{\tau} f(t) v(t)dt\right)^{q}
u(x)dx= \int_{\tau}^{b} u(x)dx \left(\int_{a}^{\tau} f(t)
v(t)dt\right)^{q}.
\end{split}
\end{equation*}
Applying the reverse H\"{o}lder inequality with powers
 $p$ and  $p'=\frac{p}{p-1}$ to the second integral of the last
expression, we get that
%
\begin{equation*}
\begin{split}
J\geq &\int_{\tau}^{b} u(x)dx \left(\int_{a}^{\tau}
v(t)^{p'}dt\right)^{\frac{q}{p'}}\left(\int_{a}^{\tau} f(t)^{p}
dt\right)^{\frac{q}{p}}\\
\geq &\int_{\tau}^{b} u(x)dx \left(\int_{a}^{\tau}
v(t)^{p'}dt\right)^{\frac{q}{p'}}\left(\int_{a}^{b} f(t)^{p}
dt\right)^{\frac{q}{p}}.
\end{split}
\end{equation*}
Thus, we obtain that
\begin{equation}\label{eq:1.4}
\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q} u(x)dx\geq
A(\tau)^q\left(\int_{a}^{b} f(t)^{p} dt\right)^{\frac{q}{p}}.
\end{equation}
It is easy to see that the condition (\ref{eq:1.3}) is equivalent
with the validity of inequality (\ref{Re.In.1}), i.e.
(\ref{eq:1.4}). If we suppose that condition (\ref{eq:1.3}) is
satisfied, i.e. if there exists $\tau\in (a,b)$ such that
$A(\tau)>0,$ then from (\ref{eq:1.4}) we have inequality
(\ref{Re.In.1}) with $C \geq A(\tau)$. Conversely, let us suppose
that inequality (\ref{Re.In.1}) holds, which means that for positive
functions $f$ such that
%
$$
\left(\int_{a}^{b}f(t)^{p} dt\right)^{\frac{1}{p}}>0,
$$
%
the expression on the left hand side of inequality (\ref{Re.In.1})
is greater than zero, i.e.
\begin{equation}\label{eq:1.5}
\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}>0.
\end{equation}
If we define
$$
a^*=\sup\{t\in [a,b), \quad \int_a^tv^{p'}(s)ds=0\};\quad
b^*=\inf\{t\in (a,b], \quad \int_t^bu(s)ds=0\},
$$
then
$$
A(t)=0\quad \mbox{for \, all} \,\, t\in (a,b)\quad \mbox{ if\, and\,
only\, if\,}\quad b^*\leq a^*.
$$
This together with (\ref{eq:1.5}) implies that
$A(t)$ is positive  for some $t\in (a,b).$\\

\noindent From (\ref{eq:1.4}) we have that
\begin{equation*}\label{eq:1.6}
A(\tau)\leq C=\inf\limits_{f>0}\frac{\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}}{\left(\int_{a}^{b} f(t)^{p}
dt\right)^{\frac{1}{p}}}.
\end{equation*}
The right hand side of the last estimate is independent on $\tau,$
so we get that%
\begin{equation*}\label{eq:1.7}
A^*=\sup\limits_{(a,b)}A(\tau)\leq C.
\end{equation*}%
This ends the proof of i).\\

\noindent If $A^*=\infty$ then inequality (\ref{Re.In.1}) holds, since
its left hand side is infinite for functions $f$ such that
$$
\left(\int_a^b f^{p}(t)dt\right)^{\frac{1}{p}}>0,
$$
which follows from (\ref{eq:1.4}).
\end{proof}


\begin{theorem}\label{rev.the-m2}
Let $0<p<1$ and $q<0.$ Then inequality (\ref{Re.In.1}) holds for all
functions $f>0$ if and only if the following condition is satisfied:
\begin{equation*}
A_*:=\inf\limits_{(a,b)}A(t)>0.
\end{equation*}
Moreover, if $C$ is the best possible constant in (\ref{Re.In.1}),
then
$$
\left(1+\frac{p'}{q}\right)^{\frac{1}{p'}}\left(1+\frac{q}{p'}\right)^{\frac{1}{q}}\,
A_* \leq C\leq A_*.
$$
\end{theorem}

\begin{proof}

\noindent({\bf Sufficiency}) Let $\alpha\in (0, -\frac{1}{p'})$ be a
parameter  and  denote
$$
V(t):=\int_{a}^{t} v^{p'}(\tau)d\tau.
$$
For
\begin{equation*}
\begin{split}
J &:= \int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^q u(x)dx\\
&= \int_{a}^{b}\left(\int_{a}^{x} f(t)
v(t)dt\right)^{p}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q-p}
u(x)dx\\
&=\int_{a}^{b}\left(\int_{a}^{x} f(t) V^{-\alpha}(t)
V^{\alpha}(t)v(t)dt\right)^{p}\left(\int_{a}^{x} f(t)
v(t)dt\right)^{q-p} u(x)dx
\end{split}
\end{equation*}
 applying the reverse H\"{o}lder inequality with powers
$p$ and $p'=\frac{p}{p-1}$ to the integral in the first brackets, we
get,
\begin{equation*}
\begin{split}
J\geq &\int_{a}^{b}\left(\int_{a}^{x} f^p(t) V^{-\alpha p}(t)
dt\right)\left(\int_{a}^{x} V^{\alpha
p'}(t)v^{p'}(t)dt\right)^{\frac{p}{p'}}\\
&\times\left(\int_{a}^{x} f(t)
v(t)dt\right)^{q-p} u(x)dx\\
%\end{split}
%\end{equation*}
%\begin{equation*}
%\begin{split}
= &\frac{1}{(1+\alpha
p')^{\frac{p}{p'}}}\int_{a}^{b}\left[\left(\int_{a}^{x} f^p(t)
V^{-\alpha p}(t) dt\right) V^{(1+\alpha
p')\frac{p}{p'}}(x)\right]\\
&\times\left(\int_{a}^{x} f(t) v(t)dt\right)^{q-p} u(x)dx.
\end{split}
\end{equation*}
Now we again apply the reverse H\"{o}lder inequality with
powers $\frac{q}{p}$ and $\frac{q}{q-p}$ which yields%
%%
\begin{equation*}
\begin{split}
J \geq & \frac{1}{(1+\alpha
p')^{\frac{p}{p'}}}\left(\int_{a}^{b}\left(\int_{a}^{x} f^p(t)
V^{-\alpha p}(t) dt\right)^{\frac{q}{p}} V^{(1+\alpha
p')\frac{q}{p'}}(x) u(x)dx\right)^{\frac{p}{q}}\\
& \times\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^q
u(x)dx\right)^{1-\frac{p}{q}}\\
= &\frac{ J^{1-\frac{p}{q}}}{(1+\alpha
p')^{\frac{p}{p'}}}\left(\int_{a}^{b}\left(\int_{a}^{x} f^p(t)
V^{-\alpha p}(t) dt\right)^{\frac{q}{p}} V^{(1+\alpha
p')\frac{q}{p'}}(x) u(x)dx\right)^{\frac{p}{q}}.
\end{split}
\end{equation*}%
%%
The reverse Minkowski integral inequality with power
$r=\frac{q}{p}$ yields%
\begin{equation*}
\begin{split}
J&\geq \frac{ J^{1-\frac{p}{q}}}{(1+\alpha
p')^{\frac{p}{p'}}}\int_{a}^{b}f^p(t) V^{-\alpha
p}(t)\left(\int_{t}^{b}V^{(1+\alpha p')\frac{q}{p'}}(x)
u(x)dx\right)^{\frac{p}{q}}dt\\
&\geq \frac{ J^{1-\frac{p}{q}}\,\, \mathbb{A}^p_*(\alpha)}{(1+\alpha
p')^{\frac{p}{p'}}}\int_{a}^{b}f^p(t)dt,
\end{split}
\end{equation*}%
where%
$$
\mathbb{A}_*(\alpha):=\inf\limits_{(a,b)}\mathbb{A}(t, \alpha)=\inf\limits_{(a,b)}V^{-\alpha
}(t)\left(\int_{t}^{b}V^{(1+\alpha p')\frac{q}{p'}}(x)
u(x)dx\right)^{\frac{1}{q}}.
$$%
Therefore, we obtain that%
\begin{equation}\label{rev.Re.In0}
J^{\frac{1}{q}} \geq \frac{\, \mathbb{A}_*(\alpha)}{(1+\alpha
p')^{\frac{1}{p'}}}\left(\int_{a}^{b}f^p(t)dt\right)^{\frac{1}{p}}.
\end{equation}%
Now we show that%
$$
\mathbb{A}_*(\alpha)\geq C_1\, A_*,
$$%
where $C_1$ depends only on $\alpha.$ Integration by parts leads to
the estimate
\begin{equation*}
\begin{split}
J_1(t,\alpha):=&\int_{t}^{b}V^{(1+\alpha p')\frac{q}{p'}}(x) u(x)dx\\
=&\int_{t}^{b}V^{(1+\alpha p')\frac{q}{p'}}(x)
d\left(-\int_x^bu(s)ds\right)\\
=&V^{(1+\alpha p')\frac{q}{p'}}(t)
\int_t^bu(s)ds-\lim\limits_{x\rightarrow b-}V^{(1+\alpha
p')\frac{q}{p'}}(x) \int_x^bu(s)ds\\
&+\frac{(1+\alpha
p')q}{p'}\int_{t}^{b}\left(\int_x^bu(s)ds\right)V^{(1+\alpha
p')\frac{q}{p'}-1}(x) dV(x)\\
\leq & V^{\alpha q}(t) A^q(t) +\frac{(1+\alpha
p')q}{p'}\int_{t}^{b}A^q(x)V^{\alpha q -1}(x) dV(x)\\
\leq & A_*^q\left[ V^{\alpha q}(t) +\frac{(1+\alpha
p')q}{p'}\int_{t}^{b}V^{\alpha q -1}(x) dV(x)\right]\\
\leq & -\frac{1}{\alpha p'}A_*^q V^{\alpha q}(t).
\end{split}
\end{equation*}
Since $J_1(t,\alpha)=\mathbb{A}^q(t,\alpha) V^{\alpha q}(t)$ due to the definition of
$\mathbb{A}(t,\alpha)$, we finally obtain that
$$
\mathbb{A}(t,\alpha)\geq (-\alpha p')^{-\frac{1}{q}}A_*,
$$
i.e.
$$
\mathbb{A}_*(\alpha)\geq (-\alpha p')^{-\frac{1}{q}}A_*,
$$
 and from (\ref{rev.Re.In0}) it follows that
$$
J^{\frac{1}{q}}\geq \frac{ (-\alpha p')^{-\frac{1}{q}}}{(1+\alpha
p')^{\frac{1}{p'}}}\, A_*\,
\left(\int_{a}^{b}f^p(t)dt\right)^{\frac{1}{p}}.
$$
For the best constant $C$ we have
$$
\sup_{\alpha\in(0,\,-\frac{1}{p'})}\frac{ (-\alpha
p')^{-\frac{1}{q}}}{(1+\alpha p')^{\frac{1}{p'}}} A_* = \left(1+\frac{p'}{q}\right)^{\frac{1}{q}}\left(1+\frac{q}{p'}\right)^{\frac{1}{q}}
A_*\leq C.
$$
The sufficiency part is proved.\\


\noindent ({\bf Necessity}) From inequality (\ref{Re.In.1}) we get
that
\begin{equation*}\label{rev.Re.In.11}
\begin{split}
C\leq & \left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^{q}
u(x)dx\right)^{\frac{1}{q}}\left(\int_{a}^{b}
f(t)^{p}dt\right)^{-\frac{1}{p}}\\
\leq & \left(\int_{\tau}^{b}\left(\int_{a}^{x}
f(t)^{p}dt\right)^{-\frac{q}{p}}\left(\int_{a}^{x} f(t)
v(t)dt\right)^{q} u(x)dx\right)^{\frac{1}{q}}.
\end{split}
\end{equation*}
If we choose
$$
f(t)=v^{p'-1}(t),
$$
then we get
\begin{equation*}
\begin{split}
C \leq & \left(\int_{\tau}^{b}\left(\int_{a}^{x}
v(t)^{p'}dt\right)^{\frac{q}{p'}} u(x)dx\right)^{\frac{1}{q}}\\
\leq & \left(\int_{a}^{\tau}
v(t)^{p'}dt\right)^{\frac{1}{p'}}\left(\int_{\tau}^{b}
u(x)dx\right)^{\frac{1}{q}}=A(\tau),
\end{split}
\end{equation*}
and consequently,
$$
A(\tau)\geq C,
$$
which proves the necessity of the condition.
\end{proof}

\begin{proposition}
Let the assumptions of Theorem \ref{rev.the-m2} be satisfied. Then the best constant $C$ of inequality (\ref{Re.In.1})
satisfies
$$
\sup\limits_{\alpha\in (0, -\frac{1}{p'})}\frac{\, \mathbb{A}_*(\alpha)}{(1+\alpha
p')^{\frac{1}{p'}}}\leq C.
$$
\end{proposition}
\begin{proof}
The proof follows from (\ref{rev.Re.In0}).
\end{proof}
\vspace{2mm}



\noindent Let us denote
\begin{equation*}\label{eq:1.9}
B(t):=\begin{cases}\Big(\mathop{supess}\limits_{(a,\,t)}v(x)\Big)\left(\int_a^tu(x)dx\right)^{\frac{1}{q}}&if \quad p=1,\\ \\
\left(\int_a^tv^{p'}(x)dx\right)^{\frac{1}{p'}}\left(\int_a^tu(x)dx\right)^{\frac{1}{q}}&if \quad p>1.
\end{cases}
\end{equation*}
Then we can formulate the following theorem:
\begin{theorem}\label{the-m2}
Let $p\geq1$ and $q<0.$ Then inequality (\ref{In.1}) holds if and only
if there exists $\tau\in (a,b)$ such that
\begin{equation*}
B(\tau)<\infty.
\end{equation*}
Moreover,
\begin{itemize}
\item[i)]\, if
$$
0< B:=\inf\limits_{(a,b)} B(t)<\infty,
$$
and $C$ is  the best constant  of inequality (\ref{In.1}) then $C\leq B;$
\item[ii)]\, if
$$
B=0
$$
then the best constant of the inequality does not exist, more
precisely, the left hand side of (\ref{In.1}) is zero for all
nonnegative  functions $f$.
\end{itemize}
\end{theorem}

\begin{proof}
Let $\tau\in (a,b)$ be arbitrary. Then
\begin{equation*}
\begin{split}
J:=&\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^q u(x)dx\geq
\int_0^{\tau}\left(\int_{a}^{x} f(t) v(t)dt\right)^q u(x)dx\\
\geq & \int_0^{\tau}\left(\int_{a}^{\tau} f(t)
v(t)dt\right)^qu(x)dx= \int_0^{\tau} u(x)dx \left(\int_{a}^{\tau}
f(t) v(t)dt\right)^q.
\end{split}
\end{equation*}
We estimate the second integral in the last expression as follows:
\begin{itemize}
\item[] If $p=1$ then
$$
\int_{a}^{\tau}
f(t) v(t)dt\leq \mathop{supess}\limits_{(a,\,\tau)} v(t) \int_{a}^{\tau}
f(t)dt.
$$
\item[] If $p>1$ then we apply the  H\"{o}lder inequality
\begin{equation*}
\int_{a}^{\tau}
f(t) v(t)dt\leq \left(\int_{a}^{\tau}
v(t)^{p'}dt\right)^{\frac{1}{p'}}\left(\int_{a}^{\tau} f(t)^{p}
dt\right)^{\frac{1}{p}}.
\end{equation*}
\end{itemize}
Consequently, we have  that
$$
\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^q u(x)dx\geq
B(\tau)^q\left(\int_{a}^{b} f(t)^{p} dt\right)^{\frac{q}{p}},
$$
i.e.
$$
\left(\int_{a}^{b}\left(\int_{a}^{x} f(t) v(t)dt\right)^q
u(x)dx\right)^{\frac{1}{q}}\leq B(\tau)\left(\int_{a}^{b} f(t)^{p}
dt\right)^{\frac{1}{p}}.
$$
The rest of the proof follows analogously as in the proof of
Theorem \ref{the-m1}.
\end{proof}

\begin{remark}
\begingroup\rm
In Theorem \ref{rev.the-m2} we supposed that
$f>0,$ which is important, since we can construct a  nonnegative function $f$ for which inequality (\ref{Re.In.1})
does not hold.
\endgroup
\end{remark}

\begin{remark}\label{remark }
\begingroup\rm
If we  denote
\begin{equation*}
A(t):=\Big(\int_t^bv^{p'}(x)dx\Big)^{\frac{1}{p'}}\Big(\int_a^tu(x)dx\Big)^{\frac{1}{q}}
\end{equation*}
and
\begin{equation*}
B(t):=\begin{cases}\Big(\mathop{supess}\limits_{(t,\,b)}v(x)\Big)\left(\int_t^bu(x)dx\right)^{\frac{1}{q}}&if \quad p=1,\\ \\
\left(\int_t^bv^{p'}(x)dx\right)^{\frac{1}{p'}}\left(\int_t^bu(x)dx\right)^{\frac{1}{q}}&if \quad p>1
\end{cases}
\end{equation*}
then we are able to formulate results analogous to  Theorems \ref{the-m1},
\ref{rev.the-m2} and  \ref{the-m2} for
inequalities (\ref{Re.In.13}) and (\ref{Re.In.12}). The formulation and the proofs are left to the reader.
\endgroup
\end{remark}

\noindent {\bf Acknowledgement.} The first author was supported by
the Institutional Research Plan No. AV0Z10190503, the other
authors were supported by the Research Plan MSM4977751301 of the
Ministry of Education, Youth and Sports of the Czech Republic.


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