# Relaxed/Truncated Version of Wiener's Tauberian Theorem

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

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

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.

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?

## 1 Answer

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 $$T\rightarrow \infty$$.