The product of $\phi$ is a Schwartz function with $m \in L^\infty$.

Question

If $\phi$ is a Schwartz function and $m \in L^\infty$, if the product $m\cdot\phi$ a Schwartz function?

The Hilbert transform is given by a multiplication with the $L^\infty$ function $m(\xi) = -i\,\text{sgn}(\xi)$, ie $$Hf(x) = (\hat{f}(\xi)m(\xi))^\vee(x), \forall \; f \in \mathcal{S}(\mathbb{R})$$

Is it true that $Hf \in \mathcal{S}(\mathbb{R})$??