Do Fourier transform on $L^1(\mathbb{R}^n)$ belong to the Schwarz space $\mathcal{S}(\mathbb{R}^n)$?

Question

Let $g\in L^1(\mathbb{R}^n)$. Is the Fourier transform of $g$

$$\overset{\wedge}{g}(y):=(2\pi)^{-n/2}\int_\mathbb{R^n} e^{i\langle y,x\rangle}g(x)dx$$

or the inverse Fourier transform of $g$

$$\overset{\vee}{g}(x):=(2\pi)^{-n/2}\int_\mathbb{R^n} e^{-i\langle y,x\rangle}g(y)dy$$

a Schwartz function?

$f$ is called a Schwartz function when $f\in C^\infty(\mathbb{R}^n)$ and $x^\alpha \partial^\beta f \in L^\infty(\mathbb{R}^n)$ for all $\alpha, \beta \in \mathbb{N}_0^n$ i.e. $f\in \mathcal{S}(\mathbb{R}^n)$.

Answer 1

Not true: The indicator function $$f(x)=\left \{\begin{split} 1 & \text{ if } |x|< 1\\ 0 & \text{ if } |x|\geq 1 \end{split}\right.$$ is $L^1$ but has the sinc function as its Fourier transform, which is not Schwartz.