Bourbaki

Proposition 4

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

Théories spectrales, chapter IV, § 2. Endomorphismes normaux

Let g∈L∞(X,μ)g\in \mathscr{L}^{\infty}(X, \mu). Denote by g~\widetilde{g} the class of gg in L∞(X,μ)L^{\infty}(X, \mu) and by S the spectrum of g~\widetilde{g}. The mapping f↦f∘gf\mapsto f\circ g (remark, p. 184) is the functional calculus mapping of C(S)\mathscr{C}(S) into L∞(X,μ)L^{\infty}(X, \mu) associated with g~\widetilde{g}.

Let f∈C(S)f\in \mathscr{C}(S). The function h=f∘gh=f\circ g is μ\mu-measurable (Lemma 5 of IV, p. 184) and bounded. Denote by f∘~gf\widetilde{\circ}g its class in L∞(X,μ)L^{\infty}(X, \mu). The mapping f↦f∘~gf\mapsto f\widetilde{\circ}g is a continuous morphism of involutive algebras with unit of C(S)\mathscr{C}(S) into L∞(X,μ)L^{\infty}(X, \mu). Since g(x)g(x) belongs to S locally μ\mu-almost everywhere (Lemma 1 of IV, p. 181 and Lemma 6), this morphism associates with the identity mapping of S the class g~\widetilde{g} of the function gg. It therefore coincides with the functional calculus mapping of g~\widetilde{g} (Prop. 7 of I, p. 111).

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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