What smoothness does integrability of Fourier transform imply?

Question

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 $f$.

Answer 1

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.

Answer 2

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 $\hat{f} \in L^1(\mathbb{R})$.