Let $\hat{g}=f.$ Is $g$ continuous?

Question

Let $f: \mathbb R \to \mathbb C$ such that there exists $g\in L^{1}(\mathbb R)$ with $\hat{g}=f.$ Then by Riemann-Lebesgue Lemma, we have $f$ is continuous and vanishing at infinity.

My Question is: Can we say $g$ is also continuous on $\mathbb R$?

Answer 1

No, we can't say that $g$ is continuous.

For example, take $g(x)=0$ for every $x \neq 0$ and $g(0)=1$. Then $g \in L^1$, $f=\hat{g}=0$, but $g$ is not continuous.

Another, less trivial, example is given by $g=\chi_{[0,1]}$. Indeed we have $f(\xi)=\hat{g}(\xi)=\frac{1 - e^{-2\pi i \xi}}{2\pi i\xi}$ (see here for a computation) which is continuous (more precisely, can be extended continuously through the origin) but $g$ is not.