Background/Definitions

Let $U_t$ be the (translation) $C_0$-group on $L^1(\mathbb{R})$ defined by $$ U_t(f)(x)\triangleq f(x-t) \, (\forall t \in \mathbb{R})(m-a.e. \, x \in \mathbb{R}). $$

The Wiener Tauberien theorem states that, for any $f \in L^1(\mathbb{R})$ if its Fourier transform $\hat{f}$ does not vanish on $\mathbb{R}$ then $$ cl \left( X_N \right):= cl\left( \left\{ U_t f: \, t \in \mathbb{R} \right\} \right) = L^1(\mathbb{R}), $$ where $cl(\cdot)$ is the closure of a set in the norm topology on $L^1(\mathbb{R})$.

Question

Is there a "truncated version" of the Wiener theorem, which gives (reasonable) sufficient condition on $f \in L^1(\mathbb{R})$ so that, for a fixed $N\in \mathbb{Z}^+$, the set $$ cl\left( \left\{ \sum_{i=1}^N \beta_i U_t(f):\, t \in \mathbb{R},\, \beta_1,\dots,\beta_N \in \mathbb{R} \right\} \right) = L^1(\mathbb{R})? $$

The motivation for my question comes as follows:

Motivation:

In the operator-Theoretic proof of J. Neerven, can be summarized as follows:

• Define the Banach algebra homomorphism $L^1(\mathbb{R})\rightarrow \mathcal{L}(L^1(\mathbb{R}))$ by $$ U(f)g\triangleq \int_{-\infty}^{\infty} f(t) U(t)(g)(x) dt = f \star g, $$ (this is the convolution operator)
• Define $X\triangleq cl(\operatorname{span}\{U_t f: \, t \in \mathbb{R}\})$ and write $U^Y$ to be the quotient operator on $L^1(\mathbb{R})/X$,
• Show that $f \in Sp(U^Y)$ where $Sp(U^Y)$ is the Averson spectrum of $U^Y$ defined by $$ \left\{ \xi \in \mathbb{R}:\, \hat{g}(\xi)=0\, \forall g \in U^Y \mbox{ with } U^Y(g)=0 \right\}, $$
• Use the fact that $Sp(U^Y)=0$ only if $U^Y=\{0\}$ to conclude that $Y=\{0\}$ since $f \in Sp(U^Y)$ and $\hat{f}$ was assumed to have no roots.

In this proof, it's not clear to me where the span operator was used and why $X$ can't be replaced by $X_N$ (as defined above)?