Relaxed/Truncated Version of Wiener's Tauberian Theorem

Question

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 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 $L^1(\mathbb{R})$.

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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 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 $d(x,z)\leq \epsilon$.

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The motivation for my question comes as follows:

Classical Proof of Wiener's Theorem:

In the operator-Theoretic proof of J. van 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.

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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?

Answer 1

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 $T\rightarrow \infty$.