Bourbaki

Lemma 9

Tag 02Z8, permanent. Cite it as [Bourbaki, Tag 02Z8].

Théories spectrales, chapter IV, § 2. Endomorphismes normaux

Let A be a commutative unital stellar subalgebra of L(E)\mathscr{L}(E). For every xx and yy in E, the mapping μ\mu of X(A)\mathsf{X}(A) into C\mathbf{C} defined by μ(f)=⟨x∣HA(f)y⟩\mu(f) =\langle x|\mathscr{H}_A(f)y\rangle for every f∈C(X(A))f\in \mathscr{C}(\mathsf{X}(A)) is a bounded measure on the compact space X(A)\mathsf{X}(A). It is positive if x=yx=y. Its norm is ⩽∥x∥∥y∥\leqslant \|x\| \|y\|, with equality if x=yx=y.

The mapping μ\mu is linear. For every function f∈C(Xf\in \mathscr{C}(\mathsf{X}(A)), one has

∣μ(f)∣⩽∥x∥∥y∥∥HA(f)∥=∥x∥∥y∥∥f∥ |\mu(f)|\leqslant \|x\| \|y\| \|\mathscr{H}_A(f)\|=\|x\| \|y\| \|f\|

hence μ\mu is a bounded measure on X(A)\mathsf{X}(A) of norm ⩽∥x∥∥y∥\leqslant \|x\| \|y\|.

Suppose that x=yx=y. For every positive function f∈C(Xf\in \mathscr{C}(\mathsf{X}(A)), the element HA(f)\mathscr{H}_A(f) is a positive element of A (example 1 of I, p. 115), hence a positive element of L\mathscr{L} (E), whence μ(f)=⟨x∣HA(f)(x)⟩⩾0\mu(f) =\langle x|\mathscr{H}_A(f)(x)\rangle \geqslant 0 (prop. 8 of I, p. 138). This shows that μ\mu is a positive measure. Since μ(1)=∥x∥2\mu(1) =\|x\|^2, the total mass of μ\mu is ∥x∥2\|x\|^2 (INT, III, p. 58, § 1, No.o8^o8, cor. 2).

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-5-6-mini, gpt-5-mini, gpt-5.4, 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.