Fourier transform of function of order $e^{-x^2}$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a measurable function such that $|f(x)|=O(e^{-x^2})$ as $|x|\to\infty$. Does it imply that $\hat{f}\in L^1(\mathbb{R})$? ($\hat{f}$ is the Fourier transform of $f$.)

( I hope this is not true but can not think of a counterexample very easily).

Answer 1

Here's a counter example. Let $f$ be the rectangular function $f(x)=1$ if $|x|<\frac 12$, and $0$ otherwise.

Then the Fourier transform is the sinc function $$f(\xi)=\frac{\sin(\pi \xi)}{\pi \xi)}$$ which is not in $L^1$.

Answer 2

Almost any such $f$ is a counterexample, because if $f,\hat f\in L^1$ then $f$ is continuous (or more carefully, then there exists a continuous function $g$ with $f=g$ almost everywhere).