What smoothness does integrability of Fourier transform imply? Suppose that $f, \hat f\in L^1(\mathbb R^n)$. What smoothness on $f$ does this imply---i.e., what is the maximal $m$ such that $f$ is $m$ times differentiable (everywhere)? What is an example of a function $f$ such that $f,\hat f\in L^1(\mathbb R^n)$ is not $m+1$ times differentiable?
All I know is that $f$ has to be continuous. If $m=0$ is the maximal $m$, this means that there exists a function that is not differentiable but whose Fourier transform is in $L^1(\mathbb R^n)$. I'm having trouble finding such an explicit function $f$.
 A: Here's an example for $n=1$.
Take $f(x) := e^{-|x|}$. Then
\begin{align*}
\int_{-\infty}^{\infty} |f(x)| \,dx &= \int_{-\infty}^{\infty} e^{-|x|} \,dx \\
&= 2 \int_0^{\infty} e^{-x} \,dx && \text{by parity of $f$}\\
&= 2 \\
&< \infty
\end{align*}
hence $f \in L^1(\mathbb{R})$. Note that $f$ is not differentiable at $0$ because its left derivative at that point is $1$ while its right derivative is $-1$.
The Fourier transform of $f$ is
\begin{align*}
\hat{f}(\xi) &= \int_{-\infty}^{\infty} e^{-|x|} e^{-i x \xi}\,dx \\
&= \int_{-\infty}^{0} e^{x(1-i\xi)} \,dx + \int_0^{\infty} e^{-x(1+i\xi)} \,dx \\
&= \frac{1}{1-i\xi} + \frac{1}{1+i\xi} \\
&= \frac{2}{1+\xi^2}
\end{align*}
and since
\begin{align*}
\int_{-\infty}^{\infty} |\hat{f}(\xi)| \,d\xi &= \int_{-\infty}^{\infty} \frac{2}{1+\xi^2} \,d\xi \\
&= 2\big(\lim_{\xi\to\infty}\arctan(\xi)-\lim_{\xi\to-\infty}\arctan(\xi)\big) \\
&= 2 \pi \\
&< \infty
\end{align*}
we have $\hat{f} \in L^1(\mathbb{R})$.
A: Consider the radial function
$$ f(r) = r e^{-r}, $$
in $\mathbb{R}^d$, which is differentiable except at zero, where it looks like $\lvert x \rvert$.
Up to constants, one can compute that the Fourier transform of this is
$$ \hat{f}(\vec{k}) = \hat{f}(k) \propto \frac{d-k^2}{(1+k^2)^{(d+3)/2}}. $$
(It turns out it's a rather unpleasant Bessel integral.)
This looks like $k^{2-d-3}=k^{-d-1}$ for large $k$, which is enough decay to be in $L^1(\mathbb{R}^d)$. Hence for any $d$, having $f$ and its Fourier transform absolutely integrable only gives continuity, not differentiability.
