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Wiener's theorem gives the necessary and sufficient conditions for the set of translates of a set of functions to be dense in $L^1(\mathbb{R}^n)$, which translates algebraically into a statement about the closed ideals of the $C^*$ algebra, $L^1(\mathbb{R}^n)$. It is known that $L^1(G)$ is a $C^*$ algebra for general groups, and that the algebra is commutative iff $G$ is. In the case $G$ is abelian, the result implies that the PNT is equivalent to $\zeta(1+it)\ne 0$ for $t\in\mathbb{R}$ after using Wiener's theorem to prove the Wiener-Ikehara theorem, and ultimately provides a very efficient proof of this result, ultimately depending on the fact that functions with non-vanishing Fourier transforms (on $\mathbb{R}^n$) are those which are not in any maximal ideal of the Fourier algebra.

In the case $G\ne\mathbb{R}^n$ but is still abelian, there is still a notion of "Fourier" transform, more appropriately called the Gelfand transform, which is a genuine generalization of the classical Fourier transform. In the case $G$ is not commutative, then this paper provides a notion of Gelfand transform for the algebra which generalizes the one for the case $G$ is abelian, now with characters of $G$ replaced by unitary representations. If one has, say, a compact group, $G$, which is not abelian, what kinds of analogous results can be obtained from a set of functions $\{f_i\}_{i\in I}\subseteq L^1(G)$ such that their generalized Gelfand transforms have no common zero? What kind of tauberian theorems come out of this, if any?

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In the non-commutative setting, what do you mean by a Gelfand transform having a zero at some point? In ${\bf C}$ non-zero = invertible but obviously this is not true in $M_n({\bf C})$ for $n\geq 2$. – Yemon Choi Apr 9 '14 at 22:57
You're right, sorry. Since the map which sends something to its Gelfand transform is basically like $x\mapsto E_x$, an evaluation map, I'm pretty sure what the right generalization is is to ask if there is no common vector in the intersection of the kernels of the Gelfand tranforms of the functions. – Adam Hughes Apr 10 '14 at 2:26

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