Bourbaki

Theorem 1

Tag 02N6, permanent. Cite it as [Bourbaki, Tag 02N6].

Théories spectrales, chapter II, § 3. Sous-espaces invariants

Let I be a closed ideal of L1(G)L^1(G) distinct from L1(G)L^1(G). There exists a character x^∈G^\widehat{x}\in \widehat{G} such that FG(f)(x^)=0\mathscr{F}_G(f)(\widehat{x}) = 0 for every f∈If\in I.

Comme the algebra L1(G)L^1(G) is regular (prop. 2 of II, p. 219) and without radical (cor. of prop. 22 of I, p. 126), and since the set of functions whose Fourier cotransform has compact support is dense in L1(G)L^1(G) (lemma 1), cor. 1 of I, p. 92 shows that the ideal I is contained in a regular maximal ideal of L1(G)L^1(G), that is to say in the kernel of a character y^\widehat{y} of L1(G)L^1(G) (th. 2 of I, p. 30); one can then take x^=y^−1(\widehat{x}=\widehat{y}^{-1}(cf. formula (8) of II, p. 207).

Corollary 1 (Wiener Tauberian Theorem)

Let f∈L1(G)f\in L^1(G). If the Fourier transform of ff does not vanish, the functions f∗εx:g↦f(gx−1)f*\varepsilon_x:g\mapsto f(gx^{-1}), where xx runs through G, form a total set in L1(G)L^1(G) (EVT, I, p. 12, def. 1).

Let V be the closed vector subspace of L1(G)L^1(G) generated by the f∗εxf*\varepsilon_x. By INT, VIII, §4, cor. of prop. 20, the space V is a closed ideal of L1(G)L^1(G). By th. 1 we have V=L1(G)V = L^1(G).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, gpt-5-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.