Intended to label questions regarding Wiener's Tauberian Theorem and related topics.
Wiener's Tauberian Theorem is a result of fundamental importance in harmonic analysis:
Theorem (WTT) Suppose $G$ is a locally compact abelian group and $I$ is a closed ideal in $L^1(G)$. If for every $\xi\in\hat G$ there exists $f\in I$ with $\hat f(\xi)\ne0$ then $I=L^1(G)$.
Apart from its intrinsic interest this is important because many, if not most or all, traditional tauberian theorems are consequences; sometimes in a very straightforward way, sometimes somewhat less so.
WTT is reminiscent of what might/should be called the fundamental theorem of commutative Banach algebras with identity:
Theorem (FTCBAID) If $I$ is a proper ideal in the cumuttative Banach algebra $A$ and $A$ has an identity then there exists a complex homomorphism $\phi$ of $A$ such that $\phi(x)=0$ for every $x\in I$.
Indeed, if $G$ is discrete then $L^1(G)$ has an identity, and WTT becomes a special case of FTCBAID. Of course in general $L^1(G)$ has no identity, so FTCBAID does not apply directly. The traditional proofs of WTT that one sees in books use Banach-algebra-ish arguments; it's satisfying to note that one can in fact derive WTT from FTCBAID (cf. Loomis Abstract Harmonic Analysis):
FTCBAID implies WTT (sketch): Suppose $K\subset\hat G$ is compact. Let $J$ be the ideal consisting of all $f\in L^1(G)$ such that $\hat f|_K=0$. Then $A=L^1/J$ is a CBA with an identity given by any $f\in L^1$ with $f|_K=1$. It's not hard to show that the maximal ideal space of $A$ "is" $K$; it follows that if $f\in L^1$ and $\hat f$ is supported in $K$ then $f\in I$. Functions with Fourier transforms with compact support are dense in $L^1$; hence $I=L^1$.