Recall that a function $f\colon \mathbb{R} \to \mathbb{C}$ is said to be rapidly decreasing if $$\sup_{x \in \mathbb{R}} \big|x^k f(x)\big| < \infty \quad \text{for all} \quad k = 0, 1, 2\dotsc$$
In the textbook [1, Lesson 19] I find the following statement:
If $f \in \mathrm{C}^{\infty}(\mathbb{R})$ is rapidly decreasing, the same is true for $\widehat{f}$.
Is this really true? If every derivative of $f$ also have rapid decay, then the result follows from the invariance of the Fourier transform on the Schwartz space $\mathscr{S}(\mathbb{R})$, but this is not assumed here.
The most general results I know of are the following:
Proposition 1. The Fourier transform of a rapidly decreasing, locally integrable function ${f \in \mathrm{L}_{\mathrm{loc}}^{1}(\mathbb{R})}$ is in $\mathrm{C}^{\infty}(\mathbb{R})$. In particular, if $f \in \mathrm{C}^{\infty}(\mathbb{R})$ is rapidly decreasing, then ${\widehat{f} \in \mathrm{C}^{\infty}(\mathbb{R})}$.
Proposition 2. Suppose that ${f \in \mathrm{C}^{\infty}(\mathbb{R}) \cap \mathrm{L}^{1}(\mathbb{R})}$ with all its derivatives ${f^{(k)}}$ in ${\mathrm{L}^{1}(\mathbb{R})}$ for ${k \in \mathbb{N}}$. Then the Fourier transform ${\widehat{f}}$ decays rapidly.
Rapid decay of the derivatives of $f$ would imply integrability of the derivatives, and hence that ${\widehat{f}}$ decays rapidly.
Source:
[1] C. Gasquet. P. Witomski. Fourier Analysis and Applications. Filtering, Numerical Computation, Wavelets.