A doubt. Let $f,g$ non negative functions on $\mathbb{R}^n$. Let $F$, $F^{-1}$ Fourier transform, Inverse Fourier transform respectively. Let $u$ a function. The following is true? If $f\leq g$ functions then $F^{-1}f(\xi)Fu(\xi)\leq F^{-1}g(\xi)Fu(\xi)$ (or $||F^{-1}f(\xi)Fu(\xi)||\leq ||F^{-1}g(\xi)Fu(\xi)||$)
I ask it to know if $F^{-1}g(\cdot )Fu\in L^p(\mathbb{R}^n)$ then $F^{-1}f(\cdot )Fu\in L^p(\mathbb{R}^n)$ with $u\in L^p(\mathbb{R}^n)$?
I have this: \begin{align*} {\left\|{\mathcal{F}}^{-1}(1+a(|x|^2))\mathcal{F} u(x)\right\|}_{L^p(\mathbb{R}^n)}^{p} &={\left\|(1+a(|x|^2))\mathcal{F}u(x)\right\|}_{L^p(\mathbb{R}^n)}^{p}\\ &=\int_{\mathbb{R}^n} |(1+a(|x|^2)\mathcal{F}u(x)|^p dx\\ &=\int_{\mathbb{R}^n} |(1+a(|x|^2)|^p|\mathcal{F}u(x)|^p dx\\ & \leq \int_{\mathbb{R}^n} |(1+b(|x|^2)|^p|\mathcal{F}u(x)|^p dx\\ &=\int_{\mathbb{R}^n} |(1+b(|x|^2)\mathcal{F}u(x)|^p dx\\ &=\left\|(1+b(|x|^2)\mathcal{F}u(x)\right\|_{L^p(\mathbb{R}^n)}\\ &=\left\|\mathcal{F}^{-1}(1+b(|x|^2)\mathcal{F}u(x)\right\|_{L^p(\mathbb{R}^n}^{p} \end{align*} (The above is wrong. Well, I thought that the Plancherel Theorem was also valid for $L^p$ but it is not ...)
Therefore $F^{-1}g(\cdot )Fu\in L^p(\mathbb{R}^n)\rightarrow F^{-1}fg(\cdot )Fu\in L^p(\mathbb{R}^n)$ ?