Bourbaki

Theorem 1 (“Spectral Theorem”)

Tag 035H, permanent. Cite it as [Bourbaki, Tag 035H].

Théories spectrales, chapter IV, § 5. Opérateurs partiels normaux et théorème spectral

Let uu be a normal partial operator on the Hilbert space E. There exist a locally compact topological space X, a positive measure μ\mu on X, an isometric isomorphism θ\theta of L2(X,μ)L^2(X, \mu) onto E, and a continuous function gg on X, such that u=θ∘mg∘θ−1u=\theta \circ m_g\circ \theta^{-1}.

The endomorphism b(u)b(u) of E is normal by definition. According to the spectral theorem for normal endomorphisms of E (Corollary 1 of IV, p. 193), there therefore exist a locally compact topological space X~\widetilde{X}, a positive measure μ\mu on X~\widetilde{X}, an isometric isomorphism θ~\widetilde{\theta} of L2(X~,μ)L^2(\widetilde{X}, \mu) onto E, and a bounded continuous function hh on X~\widetilde{X}, such that b(u)b(u) coincides with θ~∘mh∘θ~−1\widetilde{\theta}\circ m_h\circ \widetilde{\theta}^{-1}.

Let N be the closed subset of the x∈X~x\in \widetilde{X} such that ∣h(x)∣⩾1|h(x)|\geqslant 1. Since the endomorphism 1−b(u)∗b(u)=θ~∘m1−∣h∣2∘θ~−11-b(u)^*b(u) =\widetilde{\theta}\circ m_{1-|h|^2}\circ \widetilde{\theta}^{-1} is positive and injective, the set N is locally μ\mu-negligible (Lemma 7 of IV, p. 186 and cor. of Prop. 6 of IV, p. 187). Put then X=X~−X =\widetilde{X}- N. It is a locally compact space, and the restriction of functions to X induces an isometric isomorphism of L2(X~,μ)L^2(\widetilde{X}, \mu) onto L2(X,μ∣X)L^2(X, \mu|X) (Prop. 1 of IV, p. 179). The composition of θ~\widetilde{\theta} and of the restriction of functions to X therefore induces an isometric isomorphism θ\theta of L2(X,μ∣X)L^2(X, \mu|X) onto E. As b(u)=θ∘mh∣X∘θ−1b(u) =\theta \circ m_{h|X}\circ \theta^{-1}, it follows that u=θ∘mβ−1∘(h∣X)∘θ−1u=\theta \circ m_{\beta^{-1}\circ(h|X)}\circ \theta^{-1} (Lemma 2). The theorem follows from this formula.

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, 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.