$\|(g\widehat{(f|f|^{2})})^{\vee}\|_{L^{2}} \leq C \|f\|_{L^{2}}^{r} \|(g\hat{f})^{\vee}\|_{L^{2}}$ for some $r\geq 1$?

Question

Let $g\in C_{c}^{\infty} (\mathbb R)$, and $f, |f|^{2}f\in L^{2}(\mathbb R)\cap C_{0}(\mathbb R)$ (where $C_{c}(\mathbb R)$ is the class of smooth functions with compact support and $C_{0}(\mathbb R)$ is the class of continuous functions vanishing at infinity).

My Question : Can we expect: $\|(g\widehat{(f|f|^{2})})^{\vee}\|_{L^{2}} \leq C \|f\|_{L^{2}}^{r} \|(g\hat{f})^{\vee}\|_{L^{2}}$ for some $r\geq 1$? (where $C$ is some constant)

[where $\widehat{f|f|^{2}}$ denotes the Fourier transform of $f|f|^{2}$ and $(g\widehat{(f|f|^{2})})^{\vee}$ denotes the inverse Fourier transform of $g\widehat{(f|f|^{2}})$; and $|f|^{2}= f \bar{f}$]

Thanks,

EDIT: (Where is the mistake in the below argument)

Suppose $g$ has a compact support, say $K\subset \mathbb R.$ And additionally I am assuming that $g\leq 1.$ Now, \begin{eqnarray} \|(g\widehat{|f|^{2}f})^{\vee}\|_{L^{2}(\mathbb R)} & = & \|g\widehat{(|f|^{2}f)}\|_{L^{2}(\mathbb R)} \nonumber \\ & = & \|g\widehat{|f|} \ast g\widehat{|f|}\ast g\hat{f}\|_{L^{2}} \nonumber \\ & \leq & \|g\widehat{|f|}\|_{L^{1}(\mathbb R) }^{2} \|g\hat{f}\|_{L^{2}(\mathbb R)}\nonumber\\ & = & \|g\widehat{|f|}\|_{L^{1}(K)} \|g\hat{f}\|_{L^{2}}\nonumber\\ & \leq & \|g\widehat{|f|}\|_{L^{2}(K)}^{2} \|g\hat{f}\|_{L^{2}}\nonumber\\ & \leq & \|\widehat{|f|}\|^{2}_{L^{2}(K)}\|g\hat{f}\|_{L^{2}(\mathbb R)}\nonumber\\ & \leq & \|\widehat{|f|}\|^{2}_{L^{2}(\mathbb R)} \|g\hat{f}\|_{L^{2}(\mathbb R)}\nonumber\\ & = & \|f\|^{2}_{L^{2}(\mathbb R)} \|(g\hat{f})^{\vee}\|_{L^{2}(\mathbb R)}; \end{eqnarray} [In the first equality I have used Planchrel's theorem; fourth equality is due to $g$ has a compact support; fifth inequality is due the inclusion $L^{2}(K)\subset L^{1}(K)$; sixth inequality is due to $g\leq 1$; seventh inequality is due to $A\subset B \implies \int_{A} |h| dm \leq \int_{B} |h| dm$ ($dm$ Lebsgue measure)]

Answer 1

As written: no.

The scaling is not right.

Let $f_\lambda(x) = f(\lambda x)$, we have $\widehat{f_\lambda}(\xi) = \frac{1}{\lambda} \hat{f}(x / \lambda)$ and similarly $g_{\lambda}$.

Consider your inequality with $f_\lambda$ and $g_{1/\lambda}$

Using Plancherel we want to control $$ \| g_{1/\lambda} (\widehat{f_\lambda |f_\lambda|^2}) \|_{L^2}^2 = \int \left|g(x/\lambda) \frac{1}{\lambda} \widehat{f|f|^2}(x/\lambda)\right|^2 \mathrm{d}x \\ = \frac1{\lambda} \int | g(x/\lambda) \widehat{f|f|^2}(x/\lambda) | \mathrm{d}x / \lambda = \frac{1}{\lambda} \| g (\widehat{f|f|^2})\|_{L^2}^2 $$

The right hand side, however, scales like $$ \|f_\lambda\|_{L^2}^2 = \frac{1}{\lambda} \int |f(\lambda x)|^2 \lambda \mathrm{d}x = \frac1\lambda \|f\|_{L^2}^2 $$ and $$ \| g_{1/\lambda} \widehat{f_\lambda} \|_{L^2}^2 = \int | g(x/\lambda) \frac{1}{\lambda} \hat{f}(x/\lambda) |^2 \mathrm{d}x = \frac{1}{\lambda} \|g \widehat{f}\|_{L^2}^2 $$

So if the proposed inequality were to hold, you would have $$ \frac{1}{\lambda} \leq C \frac{1}{\lambda^r} \frac{1}{\lambda} $$ for $r \geq 1$ and for all $\lambda$, which is absurd.

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In a more general note: for inequalities between $L^p$ spaces, you cannot really do better than what Holder inequality gives you. The bound of an $L^2$ quantity on the left hand side by products of $L^2$ quantities on the right hand side, with no weights or derivative losses, should immediately set off a red flag. Without all the Fourier transform involved you would have something like $$ \|g f^3\|_{L^2} \leq \begin{cases} \|g f\|_{L^\infty} \|f\|_{L^4}^2 \\ \|g f\|_{L^2} \|f\|_{L^\infty}^2 \\ \|g f\|_{L^4} \|f\|_{L^8}^2 \end{cases} $$ and for intuition purposes you can imagine the Fourier transform as roughly interchanging $L^p$ with $L^{q}$ where $p^{-1} + q^{-1} = 1$, and you can see immediately, in a very rough sense, that the desired inequalities do not hold by scaling.

Answer 2

My previous answer addressed the issue of scaling. This answer addresses the issue of support. Basically:

There exists functions $g, f$ such that $g(\widehat{f|f|^2}) \not\equiv 0$ while $g \hat{f} \equiv 0$.

Let $\hat{f}$ be a non-negative real valued $C^\infty_c$ function, so that automatically $f, f|f|^2\in L^2 \cap C_0$. We have that $$ \widehat{f |f|^2} = \hat{f} \star (\hat{f} \star \hat{\bar{f}}) = \hat{f} \star (\hat{f} \star (\bar{\hat{f}} \circ \iota)) $$ where $h \circ \iota(x) = h(-x)$. As $\hat{f}$ is real and non-negative, we see that the support of $\widehat{f|f|^2}$ is strictly larger than the support of $\hat{f}$. Therefore there exists $g\in C^\infty_c$ such that $$ \operatorname{supp} g \subset (\operatorname{supp} \widehat{f|f|^2}) \setminus (\operatorname{supp} \hat{f}) $$

This argument here shows that

Even if you change the $L^p$ norms used in the inequality so that the scaling is no longer an issue, the desired inequality cannot possibly hold.