Inequality with decreasing rearrangement and non-decreasing function This question is a continuation of the question here.
Let $f^{*}$ be the usual decreasing rearrangement function of a measurable function $f$ on a measure space $(X, \mu)$. Let $1<p<n$ and set $$p'=\frac{pn}{n-p}.$$ Also, let $g$ be a positive, non-decreasing function on $\mathbb{R}_{+}$. Is it true that $$\left(\int_{0}^{\infty}(f^{*}(s))^{p'}(g(s))^{-p'}ds\right)^{p/p'}\leq c\int_{0}^{\infty}(f^{*}(s))^{p}(g(s))^{-p}s^{\frac{p}{p'}-1}ds$$ for some positive constant $c>0$ (possibly depending on function $g$)?
Any help is appreciated!
 A: $\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer to this question is yes.
Indeed,
let $h:=f^*/g$. Then $h$ is a nonincreasing function and the inequality in question can be rewritten as
\begin{equation*}
    \Big(\int_0^\infty h(s)^{p'}ds\Big)^{p/p'}
\le c\int_0^\infty h(s)^p s^{p/p'-1}\,ds. \tag{1}\label{1}
\end{equation*}
Note that $p'>p>0$.
Letting
\begin{equation*}
    a:=p'/p>1\quad\text{and}\quad u:=h^p,
\end{equation*}
we see that it is enough to show that
\begin{equation*}
    L\overset{\text{(?)}}\le \frac1a\,R, \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
    L:=\Big(\int_0^\infty u(s)^a\,ds\Big)^{1/a},\quad 
    R:=\int_0^\infty u(s)s^{1/a-1}\,ds,
\end{equation*}
and $u$ is a nonnegative nonincreasing function.
By approximation, without loss of generality (wlog) the function $u$ is piecewise constant, with just a finite number of discontinuities and with $u(s)=0$ for all large enough $s>0$. So, wlog
\begin{equation}
    u(s)=\int_{(s,\infty)}\mu(dt)
\end{equation}
for some finite measure $\mu$ with a finite support $S_\mu\subseteq(0,\infty)$ and all real $s\ge0$. Then
\begin{equation}
    R=\int_0^\infty ds\,s^{1/a-1}\,\int_{(s,\infty)}\mu(dt)
    =\int_{(0,\infty)}\mu(dt)\,\int_0^t ds\,s^{1/a-1}
    =a\int_{(0,\infty)}\mu(dt)\,t^{1/a}
        =a\int_{(0,\infty)}\nu(dt),
\end{equation}
where $\nu(dt):=\mu(dt)\,t^{1/a}$, and
\begin{equation*}
    L:=\Big(\int_0^\infty ds\,\Big(\int_{(s,\infty)}\nu(dt)\,t^{-1/a}\Big)^a\Big)^{1/a}. 
\end{equation*}
Since $a>1$, $L^a$ is convex in $\nu$ (actually, by Minkowski's inequality, even $L$ itself is convex in $\nu$), whereas $R$ is affine in $\nu$.
Note that the support of the measure $\nu$ is finite and,
by homogeneity, wlog $\nu$ is a probability measure.
So, we have the following:
Given any value of $R$, the maximum of $L$ over all probability measures $\nu$ with support $S_\nu$ in a given compact interval $I$ (say of the form $[0,N]$) and with the cardinality of $S_\nu$ not exceeding a given natural number is attained at a Dirac measure $\de_z$ supported on a singleton set $\{z\}\subseteq I$. So, wlog $\nu=\de_z$, and then
\begin{equation}
    L^a=\int_0^\infty ds\,\Big(\int_{(s,\infty)}\de_z(dt)\,t^{-1/a}\Big)^a
    =\int_0^\infty ds\,z^{-1}\,1(z>s)=1
\end{equation}
and
\begin{equation}
    R=a\int_{(0,\infty)}\de_z(dt)=a. 
\end{equation}
Thus, \eqref{1} is proved.
$\quad\Box$
We also see that the constant factor $\frac1a$ in \eqref{2} is the best possible one. So, the best possible constant factor in \eqref{1} is $c=\frac1a=\frac p{p'}$.
