**Background**
--------------


Let $(U\_t)\_{t \in \mathbb{R}}$ be the (translation) $C\_0$-group on $L^1(\mathbb{R})$ defined by
$$
U\_t(f)(x) = f(x-t) \quad \text{for almost every } x \in \mathbb{R}
$$
(for $t \in \mathbb{R}$ and $f \in L^1(\mathbb{R})$).


The [Wiener Tauberien theorem](https://en.wikipedia.org/wiki/Wiener%27s_tauberian_theorem#The_condition_in_L1) states that if $f \in L^1(\mathbb{R})$ and its Fourier transform $\hat{f}$ has no zeros on $\mathbb{R}$, then 
$$
cl\left(
\operatorname{span}\left\{
U\_t f: \, t \in \mathbb{R}
\right\}
\right)
 = L^1(\mathbb{R}),
$$
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 a (reasonable) sufficient condition on $f \in L^1(\mathbb{R})$ so that, for a fixed $N\in \mathbb{Z}^+$, there exists $\epsilon>0$ (depending on $N$) such that the set
$$
cl\left(
\left\{
\sum\_{i=1}^N \beta\_i U\_{t\_i}(f):\,
t\_1,\dots,t\_N \in \mathbb{R},\, \beta\_1,\dots,\beta\_N \in \mathbb{R}
\right\}
\right)
$$
is [$\epsilon$-dense](https://en.wikipedia.org/wiki/Dense_set#Related_notions) in $L^1(\mathbb{R})$. 


Here, a set $Z\subseteq X$ is called $\epsilon$-dense in a Banach space $X$ if for every $x \in X$ there exists some $z \in Z$ satisfying $d(x,z)\leq \epsilon$. 




---


The motivation for my question comes as follows:


**Classical Proof of Wiener's Theorem:**
----------------------------------------


In the operator-Theoretic proof of [J. van Neerven](https://fa.its.tudelft.nl/~neerven/publications/papers/RIMUT_97.pdf), 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.




---


**Edit:**
It's clear to me that this proof does not generalize to cover my question since the *span* operator cannot if $cl\left(
\operatorname{span}\left\{
U\_t f: \, t \in \mathbb{R}
\right\}
\right)$ is replaced by $cl\left\{
\sum\_{i=1}^N \beta\_i U\_{t\_i}(f):\,
t\_1,\dots,t\_N \in \mathbb{R},\, \beta\_1,\dots,\beta\_N \in \mathbb{R}
\right\}$? However, is there another approach to proving this theorem which can be modified to get around this added requirement?



Relaxed/Truncated Version of Wiener's Tauberian Theorem
No, you have the simple counter example: $$g\_{T}(x)=\frac{1}{T}1\_{0\leq x\leq T}. $$
Since $f\in L^1$, there exist $M>0$ with $\|f|\_{[-M,M]}\|\_{L^1}=\|f\|\_{L^1}-\epsilon/N$. Moreover for any $(\beta\_i,t\_i)\_{i\leq N}$ we have $$\|g\_T - \sum \beta\_i U\_{t\_i}f|\_{[-M,M]}\|\_{L^1}\geq \frac{T-2NM}{T} $$ that goes to $1$ as $T\rightarrow \infty$.


