This question is a continuation of the question [here](https://mathoverflow.net/questions/435025/inequality-with-decreasing-rearrangement-function/435042?noredirect=1#comment1120917_435042).


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!



Inequality with decreasing rearrangement and non-decreasing function
$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 $c=\frac1a=\frac p{p'}$.


