Conditions for the Fourier transform of an $L^1$ function to be in $L^1$

Question

I know that if one needs to use the Fourier inversion formula, one condition is to make sure the function $f$ and its Fourier transform $\hat{f}$ both lie in $L^1$. If now I have a function $f$ which lies in both $L^1$ and $L^2$, what kind of additional conditions can make its Fourier transform also lies in $L^1$? Thank you!

Answer 1

If $f\in C^{n+1}(\mathbb R^{n})$ and $\partial^{\alpha}f\in L^{1}\bigcap C_{0}$ for $|\alpha |\leq n+1$, then $\mid \hat{f}(\xi) \mid \leq C (1+ \mid \xi \mid)^{-n-1}$ and hence $\hat{f} \in L^{1}(\mathbb R^{n})$. [For the proof you may see, Real Analysis by G. B. Folland, Chapter 8]

Answer 2

From Suppose that $f \in L^1 $ and $\hat{f} \geq 0$. If $f$ is continuous in $0$ then $\hat{f} \in L^1$, another sufficient condition is that

[$f\in L^1, \hat f \geq 0$ and $f$ is continuous at $0$] $\implies \hat f \in L^1$.

Although the proof provided is about $f\in L^1(\mathbb R)$, it seems that due to nice properties of the Gaussian (e.g. Fourier transform on $\mathbb R^n$ of Gaussian function) the same is true for $f\in L^1(\mathbb R^n)$.

EDIT: I just learned from this answer that Bernstein gave another sufficient condition

[$f\in C_c(\mathbb R)$ and $\alpha$-Holder continuous for $\alpha>\frac 12$] $\implies \hat f \in L^1(\mathbb R)$.