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$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 bounded 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 $c=\frac1a$.

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 $c=\frac1a=\frac p{p'}$.

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 for all real $N>1$ \begin{equation*} L^{1/a}\overset{\text{(?)}}\le2R, \tag{2}\label{2} \end{equation*} where \begin{equation*} L:=\int_0^N u(s)^a\,ds,\quad R:=\int_0^N u(s)s^{1/a-1}\,ds, \end{equation*} and $u\colon[0,N]\to\R$ is a nonnegative nonincreasing function.

If $R=0$, then $L=0$ and hence inequality \eqref{2} is trivial. So, in what follows assume that $R>0$.

Then, by "vertical" rescaling, without loss of generality (wlog) $u(0)=1$.

The set, say $U_N$, of all nonnegative nonincreasing functions $u$ on the interval $[0,N]$ with $u(0)=1$ is sequentially compact in the topology of almost everywhere convergence. So, there is a maximizer of $L$ over all $u\in U_N$ with a fixed positive value of $R$.

In what follows, wlog let $u$ be such a maximizer. Let \begin{equation*} \de:=\max\{t\in[0,N]\colon u=1\text{ on }[0,t)\}, \end{equation*} \begin{equation*} T:=\max\{t\in[0,N]\colon u>0\text{ on }[0,t)\}. \end{equation*} Note that $0\le\de\le T\le N$, $u=1$ on $[0,\de)$, $0<u<1$ on $(\de,T)$, and $u=0$ on $(T,N]$.

If $\de=T$, then $L=\de$ and $R=a\de^{1/a}$, so that \eqref{2} holds trivially. So, wlog $\de<T$.

Using now Lagrange multipliers, we see that for some real $\la$ and almost all $s\in(\de,T)$ we have $u(s)^{a-1}=\la s^{1/a-1}$; it also follows that $\la>0$. So, \begin{equation*} u(s)=b s^{-1/a} \tag{5}\label{5} \end{equation*} for some real $b>0$. Wlog, \eqref{5} holds for all $s\in(\de,T)$. So, $b\de^{-1/a}=u(\de+)\le u(0)=1$ and hence \begin{equation*} b\de^{-1/a}\le1. \tag{6}\label{6} \end{equation*}

If now $\de=0$, then $R=\infty$, so that \eqref{2} holds trivially again. So, wlog $\de>0$.

Now \eqref{2} becomes \begin{equation*} (\de+b^a l)^{1/a}\le2(a\de^{1/a}+bl), \end{equation*} where $l:=\ln(T/\de)>0$. Therefore and because $a>1$, it is enough to show that \begin{equation*} \de^{1/a}+b l^{1/a}\le2(a\de^{1/a}+bl). \tag{7}\label{7} \end{equation*} If $l\ge1$, then $l^{1/a}\le l$, so that \eqref{7} follows. If $l\le1$, then $b l^{1/a}\le b\le\de^{1/a}$ by \eqref{6}, so that \eqref{7} again follows.  $\quad\Box$

Working slightly harder at the end of the above proof, we can see that \eqref{1} holds with $c=1/a<1$.

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 bounded 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 $c=\frac1a$.

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 for all real $N>1$ \begin{equation*} L^{1/a}\overset{\text{(?)}}\le2R, \tag{2}\label{2} \end{equation*} where \begin{equation*} L:=\int_0^N u(s)^a\,ds,\quad R:=\int_0^N u(s)s^{1/a-1}\,ds, \end{equation*} and $u\colon[0,N]\to\R$ is a nonnegative nonincreasing function.

If $R=0$, then $L=0$ and hence inequality \eqref{2} is trivial. So, in what follows assume that $R>0$.

Then, by "vertical" rescaling, without loss of generality (wlog) $u(0)=1$.

The set, say $U_N$, of all nonnegative nonincreasing functions $u$ on the interval $[0,N]$ with $u(0)=1$ is sequentially compact in the topology of almost everywhere convergence. So, there is a maximizer of $L$ over all $u\in U_N$ with a fixed positive value of $R$.

In what follows, wlog let $u$ be such a maximizer. Let \begin{equation*} \de:=\max\{t\in[0,N]\colon u=1\text{ on }[0,t)\}, \end{equation*} \begin{equation*} T:=\max\{t\in[0,N]\colon u>0\text{ on }[0,t)\}. \end{equation*} Note that $0\le\de\le T\le N$, $u=1$ on $[0,\de)$, $0<u<1$ on $(\de,T)$, and $u=0$ on $(T,N]$.

If $\de=T$, then $L=\de$ and $R=a\de^{1/a}$, so that \eqref{2} holds trivially. So, wlog $\de<T$.

Using now Lagrange multipliers, we see that for some real $\la$ and almost all $s\in(\de,T)$ we have $u(s)^{a-1}=\la s^{1/a-1}$; it also follows that $\la>0$. So, \begin{equation*} u(s)=b s^{-1/a} \tag{5}\label{5} \end{equation*} for some real $b>0$. Wlog, \eqref{5} holds for all $s\in(\de,T)$. So, $b\de^{-1/a}=u(\de+)\le u(0)=1$ and hence \begin{equation*} b\de^{-1/a}\le1. \tag{6}\label{6} \end{equation*}

If now $\de=0$, then $R=\infty$, so that \eqref{2} holds trivially again. So, wlog $\de>0$.

Now \eqref{2} becomes \begin{equation*} (\de+b^a l)^{1/a}\le2(a\de^{1/a}+bl), \end{equation*} where $l:=\ln(T/\de)>0$. Therefore and because $a>1$, it is enough to show that \begin{equation*} \de^{1/a}+b l^{1/a}\le2(a\de^{1/a}+bl). \tag{7}\label{7} \end{equation*} If $l\ge1$, then $l^{1/a}\le l$, so that \eqref{7} follows. If $l\le1$, then $b l^{1/a}\le b\le\de^{1/a}$ by \eqref{6}, so that \eqref{7} again follows. $\quad\Box$

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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 for all real $N>1$ \begin{equation*} L^{1/a}\overset{\text{(?)}}\le2R, \tag{2}\label{2} \end{equation*} where \begin{equation*} L:=\int_0^N u(s)^a\,ds,\quad R:=\int_0^N u(s)s^{1/a-1}\,ds, \end{equation*} and $u\colon[0,N]\to\R$ is a nonnegative nonincreasing function.

If $R=0$, then $L=0$ and hence inequality \eqref{2} is trivial. So, in what follows assume that $R>0$.

Then, by "vertical" rescaling, without loss of generality (wlog) $u(0)=1$.

The set, say $U_N$, of all nonnegative nonincreasing functions $u$ on the interval $[0,N]$ with $u(0)=1$ is sequentially compact in the topology of almost everywhere convergence. So, there is a maximizer of $L$ over all $u\in U_N$ with a fixed positive value of $R$.

In what follows, wlog let $u$ be such a maximizer. Let \begin{equation*} \de:=\max\{t\in[0,N]\colon u=1\text{ on }[0,t)\}, \end{equation*} \begin{equation*} T:=\max\{t\in[0,N]\colon u>0\text{ on }[0,t)\}. \end{equation*} Note that $0\le\de\le T\le N$, $u=1$ on $[0,\de)$, $0<u<1$ on $(\de,T)$, and $u=0$ on $(T,N]$.

If $\de=T$, then $L=\de$ and $R=a\de^{1/a}$, so that \eqref{2} holds trivially. So, wlog $\de<T$.

Using now Lagrange multipliers, we see that for some real $\la$ and almost all $s\in(\de,T)$ we have $u(s)^{a-1}=\la s^{1/a-1}$; it also follows that $\la>0$. So, \begin{equation*} u(s)=b s^{-1/a} \tag{5}\label{5} \end{equation*} for some real $b>0$. Wlog, \eqref{5} holds for all $s\in(\de,T)$. So, $b\de^{-1/a}=u(\de+)\le u(0)=1$ and hence \begin{equation*} b\de^{-1/a}\le1. \tag{6}\label{6} \end{equation*}

If now $\de=0$, then $R=\infty$, so that \eqref{2} holds trivially again. So, wlog $\de>0$.

Now \eqref{2} becomes \begin{equation*} (\de+b^a l)^{1/a}\le2(a\de^{1/a}+bl), \end{equation*} where $l:=\ln(T/\de)>0$. Therefore and because $a>1$, it is enough to show that \begin{equation*} \de^{1/a}+b l^{1/a}\le2(a\de^{1/a}+bl). \tag{7}\label{7} \end{equation*} If $l\ge1$, then $l^{1/a}\le l$, so that \eqref{7} follows. If $l\le1$, then $b l^{1/a}\le b\le\de^{1/a}$ by \eqref{6}, so that \eqref{7} again follows. $\quad\Box$

Working slightly harder at the end of the above proof, we can see that \eqref{1} holds with $c=1/a<1$.