Does the operator $(\hat{f}\cdot m )^\vee$ maps Schwartz in it self?

Question

Given $m \in L^\infty$ and $\phi \in \mathcal{S}$ a Schwartz function, is it true that $(\hat{f}\cdot m)^\vee$ is a Schwartz function??

I trying to prove this so I could conclude that operator of the form $(\hat{f}\cdot m)^\vee$ maps $\mathcal{S}$ to it self.

Attempt: Given $\alpha, \beta$ multi-index, we have to prove that $$\sup_{x \in \mathbb{R^n}}|x^\alpha\partial^\beta(\hat{f}\cdot m)^\vee(x)| < \infty. $$

When $\alpha = 0$, using some properties of Fourier tranform, we get $$\partial^\beta(\hat{f}\cdot m)^\vee(x) = ((2\pi i \xi)^\beta\hat{f}(\xi)m(\xi))^\vee(x) = ((\partial^\beta f)^\wedge\cdot m)^\vee(x).$$ Then, taking the absolute value of the expression above and by definition of inverse Fourier transform, \begin{align*} |\partial^\beta(\hat{f}\cdot m)^\vee(x)| = & \left| \int_\mathbb{R^n} (\partial^\beta f)^\wedge(\xi)m(\xi) e^{2\pi i \xi\cdot x} d\xi \right| \\ \leq & \int_\mathbb{R^n} |(\partial^\beta f)^\wedge(\xi)||m(\xi)|d\xi \\ \leq &\|m\|_{L^\infty} \int_\mathbb{R^n} |(\partial^\beta f)^\wedge(\xi)|d\xi \\ =& \|m\|_{L^\infty} \|(\partial^\beta f)^\wedge\|_{L^1}. \end{align*} The $L^1$-norm of $(\partial^\beta f)^\wedge$ is finite, since this is a Schwartz function.

My problem is for $\alpha \neq 0$. For simplicity and in view of properties of Fourier transform, I changed $x^\alpha$ for $(-2\pi i x)^\alpha$ and I want to show that the supreme of $|(-2\pi i x)^\alpha \partial^\beta(\hat{f}\cdot m)^\vee(x)|$ over all $x \in \mathbb{R^n}$ is finite:

\begin{align*} |(-2\pi i x)^\alpha \partial^\beta(\hat{f}\cdot m)^\vee(x)| = & |(-2\pi i x)^\alpha ((\partial^\beta f)^\wedge\cdot m)^\vee(x)| \\ = & |[\partial^\alpha((\partial^\beta f)^\wedge \cdot m)]^\vee(x)|. \end{align*}

How do I proceed from here?? Does it make sense the derivative $\partial^\alpha((\partial^\beta f)^\wedge \cdot m)$ ??

Answer 1

If $(\hat{f} \cdot m )^\vee$ is Schwarz, then so is $\hat{f}\cdot m$ (as the Fourier transform sends Schwarz space into itself). However, $\hat{f} \cdot m$ needs not even to be continuous (as we only assume $m\in L^\infty$).

So, no. The function $(\hat{f} \cdot m)^\vee$ needs not to be Schwarz.