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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 $\hat h(y)\ll 1/y^2$

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Your condition defines a space of functions (80% a joke). $H^2$ is optimal on the $H^s$-scale ($s \in \mathbb{R}$) so you can only do slightly better. I think a somewhat weaker condition is that the first derivative is of bounded variation ... –  Helge Aug 5 '11 at 0:12
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I recognize my condition on the transform defines a space of test functions - I was wondering if it had a name. –  Stopple Aug 5 '11 at 0:31

1 Answer 1

up vote 2 down vote accepted

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.

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