# Fourier transform of Schwartz functions

I am stuck with the following question:

Suppose $f \in \mathcal{S}(\mathbb{R})$ satisfies $\widehat{f}(\xi)=0$ for $|\xi|<1.$

Prove that there exists $g \in \mathcal{S}(\mathbb{R})$ such that $f=g''$.

Here are my thoughts so far:

If such a $g \in \mathcal{S}(\mathbb{R})$ existed, then we would have $$\widehat{f}(\xi)=-4\pi^2\xi^2\widehat{g}(\xi)$$ and so, by the Fourier inversion theorem, it suffices to prove that the function $$h(\xi)=\frac{\widehat{f}(\xi)}{-4\pi^2\xi^2}$$ (where we define $h(0)=0$) is Schwartz.

My thinking here is that any possible bad (i.e. non-Schwartz) behaviour for this function can only occur when $|\xi|<1$ because otherwise $|h(\xi)|<|\widehat{f}(\xi)|$ but $h(\xi)=0$ for $|\xi|<1$ by hypothesis. Hence $h$ is Schwartz.

I'm not sure if this is right (and even if it is, I'm finding it hard to make rigorous). Any help would be appreciated. Thanks.

If $\phi\in C^\infty (\mathbb {R}),$ and $\phi = 0$ on $[-1,1],$ then $\phi (x)/x^2 \in C^\infty (\mathbb {R}).$ Proof: Being $C^\infty$ is a local property. This function is certainly in $C^\infty (\mathbb {R}\setminus \{0\})$ and it equals the $C^\infty$ function $0$ near $0.$