Let $\mathcal F$ denote the Fourier transform on $L^1(\mathbb R)$, that is, $\mathcal Ff(y) = \langle f,e^{2\pi iy\cdot}\rangle$. I found that if $g\in\mathcal FL^1(\mathbb R)$, then there is some $F\in L^1(0,1)$ such that $\hat F(n) = g(n)$ for all $n\in\mathbb Z$. To see this, let $g = \hat f$ with $f\in L^1(\mathbb R)$ and define $F(x) = \sum_{k\in\mathbb Z}f(x+k)$. Then $F$ is well defined (a.e.) and $F\in L^1(0,1)$. Also, $\hat F(n) = \hat f(n) = g(n)$.
My question is whether the following converse statement (interpolation theorem) holds:
Given $F\in L^1(0,1)$, does there exist $g\in\mathcal FL^1(\mathbb R)$ such that $\hat F(n) = g(n)$ for $n\in\mathbb Z$?
Clearly, there exists $g\in C_0(\mathbb R)$ since $\hat F(n)\to 0$ as $|n|\to\infty$. But it is well known that $\mathcal FL^1(\mathbb R)\neq C_0(\mathbb R)$.
Extend $F$ by $0$ outside of $(0,1)$; this is a function in $L^1(\mathbb{R})$. Let $g$ be its Fourier transform; it has the property $g(n) = \hat F(n)$ by construction.