Let $f\in L^1(R)$. I'm interested in necessary and/or sufficient conditions on $f$ that guarantee its Fourier transform $\hat f$ to be in $L^1(R)$.
(I think, if $f$ has compact support and is in $C^1$, then $\hat f\in L^1$, but can't think of a proof right now; although $\hat f\in L^2$ is immediate from Plancherel's theorem.)
Here is one condition (on $\hat f$ rather than $f$) that I'm aware of that is not hard to prove:
- If $f\in L^1$ is continuous and $\hat f\ge 0$, then $\hat f\in L^1$.
One sufficient condition is the following:
If $f$ has a continuous second derivative such that both $f$ and $f''$ belong to $L^1(\mathbb{R})$ then $\hat{f}\in L^1(\mathbb{R})$.
Indeed, if $g =f-f''$ then $g \in L^1(\mathbb{R})$ and $\hat{g}(\xi)=(1+4\pi^2\xi^2)\hat{f}(\xi)$. Thus $$ |\hat{f}(\xi)|\le \frac{1}{1+4\pi^2\xi^2}\Vert g\Vert_1.$$ Consequently, $\hat{f}\in L^1(\mathbb{R})$.