Showing or refuting that $f\leq g$ implies $\left\|\mathcal{F}(f\widehat{u})\right\|_{L^p}\leq \left\|\mathcal{F}(g\widehat{u})\right\|_{L^p}$

Question

Let $L^p=L^p(\mathbb{R}^n)$.

If $f\leq g$ then $\left\|\mathcal{F}^{-1}(f\widehat{u})\right\|_{L^2}\leq \left\|\mathcal{F}^{-1}(g\widehat{u})\right\|_{L^2}$ by Plancherel theorem.

If $f\leq g$ then $\left\|\mathcal{F}^{-1}(f\widehat{u})\right\|_{L^p}\leq \left\|\mathcal{F}^{-1}(g\widehat{u})\right\|_{L^p}$ ($1<p<\infty)$?

Answer 1

Even the first inequality is false, just take $f=-1$ and $g=0$, then you would get $$ \|u\|_{L^p} \leq 0 $$ which is false except if $u=0$.