# What conditions on $f\in L^1$ guarantee that $\hat f\in L^1$?

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 $\hat f\in L^1$.
• It might be helpful to think about the smoothness-decay duality..if you have a sufficiently "rough" function $f \in L^{1}$ then $\hat{f}$ will decay sufficiently slowly in the Fourier domain so that it will not be $L^{1}$. This obviously doesn't answer your question but it might point you in the right direction. Commented Dec 13, 2016 at 18:11
• If $f,\hat{f}$ are in $L^1$, then $f,\hat{f}\in L^1\cap L^\infty$. So, for example, $|f|^p \le \|f\|_{\infty}^{p-1}|f|$ for $1 \le p < \infty$. So $f,\hat{f}$ must be in every $L^p$ space for $1 \le p \le \infty$. It may be close to sufficient to assume this for $f$ only. Commented Dec 19, 2016 at 20:10
• A necessary condition is that $f(x) \to 0$ when $|x| \to +\infty$, by Riemann-Lebesgue lemma. Commented Dec 20, 2016 at 19:09

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