Easy question about fourier transform I think this is an easy one ... but I just can't find an answer.
Assume $f : \mathbb{R} \to \mathbb{R}$ is a $n+2$ times differentiable function and
and all drivatives up to the order $n+2$ are in $L_1$. From calculus we know 
that $\mathcal{F}(D^{n+2} f )(x) =  (ix)^{n+2}\hat{f}$. Here $\mathcal{F}$ and $\hat{\cdot}$ denote the Fourier-transform. By the Riemann-Lebesgue lemma we have $(ix)^{n+2}\mathcal{F}(f) \to 0$ for $|x| \to \infty$. Thus by integration we can show that $x \mapsto (ix)^{n}\hat{f}(x)$ is a $L_1$ function.
Question: I want to know if there are weaker conditions on the function $f$ that imply 
$\int_{-\infty}^{\infty} |x|^{n}|\hat{f}(x)| dx < \infty$.
Thank you !
 A: Let's deal with $n=0$ case; the general case amounts to applying $n=0$ to the $n$th derivative of $f$. The necessary and sufficient condition would be "$f$ is the inverse Fourier transform of an $L^1$ function", which of course isn't very helpful: there is no explicit description of the image of $L^1$ under the inverse (or direct) Fourier transform. At least we see that $f$ must be in $C_0$: this is necessary, but not sufficient. 
A sufficient condition is: there is $p\in (1,2]$ such that  $f,f' \in L^p(\mathbb R)$. Here you can interpret $f'$ as the pointwise derivative or as the $L^p$ derivative, i.e., the limit of $(f(\cdot +h)-f)/h$ in $L^p$: see $L^p$ derivative vs normal derivative.
Proof: the Hausdorff-Young inequality tells you that $\hat f$ and $\xi \hat f(\xi)$ are in $L^q$, where $q\in [2,\infty)$ is conjugate to $p$. By Hölder's inequality 
$$ \begin{split}
\int_{\mathbb R} |\hat f(\xi)|\,d\xi &= \int_{\mathbb R} (1+|\xi|)^{-1} (1+|\xi|) |\hat f(\xi)|\,d\xi \\
& \le \left(\int_{\mathbb R} (1+|\xi|)^{-p}\,d\xi\right)^{1/p}
\left(\int_{\mathbb R} (1+|\xi|)^{q} |\hat f(\xi)|^q\,d\xi\right)^{1/q} <\infty
 \end{split} $$
