For an application to analytic number theory, I'm considering the Fourier transform of test functions $h(x)$ on $\mathbb R$ with compact support, so the transforms $\hat h(y)$ are smooth. But I want to add a mild condition on $h$ that will force $\hat h(y) \ll 1/y^2$ as $y\to\infty$. Based on the examples I know, it looks like "continuous and piecewise differentiable" works, but I'd like the largest space of test functions possible. I'm familiar with the definition of $L^2$-based Sobolev Spaces $H^k(\mathbb R)$, but $k=1$ seems too weak, and $k=2$ too strong.



> 
> What is the right space of test functions $h(x)$ so that $\hat h(y)\ll 1/y^2$
> 
> 
> 



Fourier Transform: Smoothness and Decay  
Your space is some sort of Besov space, 
containing $B^{2+\epsilon}\_{1,\infty}$.


Using a Littlewood-Paley decomposition $1=\sum\_\nu\varphi\_\nu(\xi)$
with $\varphi\_0\in C^\infty\_c$, $\varphi\_\nu(\xi)=\varphi (\xi 2^{-\nu})$, 
$\varphi\in C^\infty\_c$ supported in a ring
$1/2\le\vert\eta\vert\le 2$, a fonction $u$ belongs to your space means
$$
\lim\_{\nu\rightarrow+\infty}2^{2\nu}\Vert\varphi\_\nu(D) u\Vert\_{\mathcal FL^\infty}=0
$$
whereas
$B^{2+\epsilon}\_{1,\infty}$ is defined by


$$
\sup\_{\nu}2^{(2+\epsilon)\nu}\Vert\varphi\_\nu(D) u\Vert\_{L^1}<+\infty
$$
Since
$$
2^{2\nu}\Vert\varphi\_\nu(D) u\Vert\_{\mathcal FL^\infty}\le
2^{2\nu}\Vert\varphi\_\nu(D) u\Vert\_{L^1},
$$
the inclusion of $B^{2+\epsilon}\_{1,\infty}$ in your space is clear.


