Inequality with decreasing rearrangement function

Question

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 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 $c>0$ (later added: possibly depending on function $g$)?

Maybe it helps that, for $p, q\in (0, +\infty)$, the Lorentz space is defined via the quasi-norm $$||f||_{p, q}=\left(\int_{0}^{\infty}(f^{*}(s))^{q}s^{\frac{q}{p}-1}ds\right)^{1/q}$$ and we have the Hoelder inequality $$||fg||_{p, q}\leq C||f||_{p_{1},q_{1}}||g||_{p_{2},q_{2}},$$ where $\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}$ and $\frac{1}{q}=\frac{1}{q_{1}}+\frac{1}{q_{2}}$.

Any help is appreciated!

In the case when $g$ is non-decreasing, see here.

Answer 1

No (if $c$ cannot depend on $f^*$ or $g$). Indeed, let $h:=f^*/g$. Then $h$ can be any positive function and the inequality in question can be rewritten as $$lhs:=\Big(\int_0^\infty h(s)^{p'}ds\Big)^{p/p'} \le rhs:=c\int_0^\infty h(s)^p s^{p/p'-1}\,ds. \tag{1}\label{1}$$ Note that $p'>p>0$.

To obtain a contradiction, suppose that \eqref{1} holds for all positive functions $h$. Then it holds for all nonnegative functions $h$. Take now any real $k>0$ and let $h:=1_{(k,k+1)}$. Then $lhs=1$ while $rhs<ck^{p/p'-1}\to0$ as $k\to\infty$. So, we get $1\le0$, a contradiction. $\quad\Box$

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The OP has changed the question, thus invalidating the above answer.

After the change, the answer is still no. Indeed, let us leverage the idea in the above answer as follows. Recall that $p'>p>0$, so that $$a:=\frac2{1-p/p'}>2.$$ Let $(k_j)_{j\ge0}$ be a strictly increasing sequence of natural numbers such that $$k_j\sim j^a$$ (as $j\to\infty$). Let $(f_n)_{n\ge0}$ be any sequence of strictly positive real numbers decreasing to $0$. Let $$f^*:=\sum_{n\ge0}f_n\,1_{[n,n+1)},\quad g:=\sum_{j\ge0}f_{k_j}\,\big(1_{[k_j,k_j+1)} +(j+1)^{(a+2)/p}\,1_{[k_j+1,k_{j+1})}\big).$$ Then $$lhs\ge\Big(\sum_{j\ge0}\int_{[k_j,k_j+1)}1\Big)^{p/p'}=\infty,$$ whereas $$rhs<c\sum_{j\ge0}\big(k_j^{p/p'-1}+(j+1)^{-(a+2)}k_{j+1}\big)<\infty,$$ since $k_j^{p/p'-1}\sim j^{-2}$ and $k_{j+1}\sim j^a$. So, in general the inequality $lhs\le rhs$ cannot hold for any real $c>0$, even if $c$ is allowed to depend on both $f^*$ and $g$. $\quad\Box$