Bourbaki

Proposition 11

Tag 02E8, permanent. Cite it as [Bourbaki, Tag 02E8].

Théories spectrales, chapter I, § 6. Algèbres stellaires

Let A be a unital involutive algebra, and let a,xa,x and yy be elements of A. Suppose that xx and yy are normal. If xa=xa= ayay, then one has f(x)a=af(y)f(x)a=af(y) for every function ff continuous in the union of the spectrum of xx and the spectrum of yy. In particular, one has f(aa∗)a=af(a∗a)f(aa^*)a=af(a^*a) for every function f∈Cf\in \mathscr{C} (Sp′(a∗a)'(a^*a)).

Let S = Sp(xx)∪\cup Sp(yy). Proposition 6 of I, p. 106 implies that x∗a=ay∗x^*a=ay^*. Consequently, one has f(x)a=af(y)f(x)a=af(y) for every function ff of the form z↦P(z,z)z\mapsto P(z, z), where P∈C[X,Y]P\in \mathbf{C}[X,Y] is a polynomial. Since the set of functions f∈C(S)f\in \mathscr{C}(S) satisfying f(x)a=f(x)a= af(y)af(y) is a closed subalgebra of C(S)\mathscr{C}(S), it coincides with C(S)\mathscr{C}(S) by TG, X, p. 40, Cor. 1.

The second assertion is a consequence of the first, applied to the hermitian elements x=aa∗x=aa^* and y=a∗ay=a^*a, having regard to the fact that Sp′(a∗a)='(a^*a) = Sp′(aa∗)'(aa^*) (Prop. 1 of I, p. 5).

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-5.4, 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.