Let $f$ be a function with
a) $f\in L^1(\mathbb R^d)$
b) $f\in \mathcal S(\mathbb R^d)$ (The Schwartz-space)
c) $f\in L^2(\mathbb R^d)$
and given the following statements:\
(i) $f$ continuous $\Rightarrow \hat{f}$ continuous
(ii) $f$ smooth $\Rightarrow \hat{f}$ smooth.\
I know (i) and (ii) are false for c) because the Fourier transform of $f(x)=\frac{\sin x}{x}$ is not continuous.
(i) and (ii) are correct because the Fourier transform of a Schwartz function is a Schwartz function and per definition a Schwartz function is smooth.
But what about a)?
As the fourier transform maps $f \in L^1(\mathbb{R}^d)$ to a continuous function, (i) is true there too.
(ii) shouldn't be true. Consider $x \mapsto \left(\frac{sin(x)}{x}\right)^2$ and its fourier transform.