Let $f \in L^1 \cap L^2$. Are there any natural conditions on $f$ that ensure that the Fourier transform $\hat f$ is in $L^1?$
I don't want to have anything as restrictive as Schwartz. I am rather looking for some very mild conditions on $f$ that ensure this.
To fix notations let us consider the Fourier transformation as acting on functions over $\mathbb R^n$. The set of functions with Fourier transform $\hat{f} \in L^1(\mathbb{R}^n)$ is the Wiener algebra. It is contained in the space of continuous functions going to zero at infinity and it contains (for example) the Sobolev space $H^{\frac{n}{2}+\epsilon}$ for any $\epsilon>0$.
In dual form, you're asking for conditions that characterize (or at least guarantee) that a given $L^1$ function is the Fourier transform of another function. To quote the introduction of Stein and Weiss' book on Fourier Analysis:
"Theorem 1.2 gives a necessary condition for a function to be a Fourier transform. Belonging to the class $C_0$, however, is far from being sufficient. There seems to be no simply satisfactory condition characterizing Fourier transforms of functions on $L^1(\Bbb R^n)$."
We have the following characterization:
The space $V:=\{\widehat f\colon f\in L^1(\textbf R)\}$ is equal to the space $W=\{h*g\colon h \text{ and } g\in L^2(\textbf R)\}$. That is the Fourier transforms of functions of $L^1(\textbf R)$ coincide with the convolutions of functions in $L^2(\textbf R)$.
Proof: Let $f\in L^1(\mathbf R)$. Then $f=uv$ with $u$ and $v\in L^2(\mathbf R)$ (for example take $u=v=\sqrt{f}$ where $\sqrt{z}$ is the unique square root with argument in $(-\pi,\pi]$). Then $$\widehat f=\widehat{uv}=\widehat u*\widehat v,$$ with $\widehat u$ and $\widehat v\in L^2(\mathbf R)$. Therefore $V\subset W$.
Let now $h$ and $g\in L^2(\mathbf R)$, since the Fourier transform is a bijective isometry in $L^2(\mathbf R)$, there are $u$ and $v\in L^2(\mathbf R)$ with $\widehat u=h$ and $\widehat v=g$, then $$h*g=\widehat u *\widehat v=\widehat {u v}$$ with $uv\in L^1(\mathbf R)$ as a product of two $L^2(\mathbf R)$ functions. Therefore $W\subset V$ End of Proof
Probably not what you want, but related.
