Theorem 5
Tag 02AS, permanent. Cite it as [Bourbaki, Tag 02AS].
Théories spectrales, chapter I, § 4. Calcul fonctionnel holomorphe
Let A be a unital Banach algebra, not necessarily commutative. Let be an element of A and the germ of the identity function of in a neighbourhood of Sp. There exists a unique continuous unital morphism of (Sp) into A such that .
The image of is contained in the full closed subalgebra of A generated by . In particular, it is contained in the bicommutant of .
Let us prove the existence of the morphism . Let B be the full closed subalgebra of A generated by . It is commutative, and we have Sp Sp (I, p. 5, No). The mapping of the holomorphic functional calculus on B is a continuous unital morphism of (Sp) into B such that (Theorem 1 of I, p. 51). The morphism composed of and the canonical injection of B into A is a continuous unital morphism of (Sp) into A such that the image of is .
Let us prove uniqueness. Let be a continuous unital morphism from (Sp) into A such that . Then and coincide on the set of germs of polynomials in a neighbourhood of Sp, hence on the set of germs of rational fractions holomorphic in a neighbourhood of Sp. But these germs are dense in (Sp) (I, p. 69, th. 3). This implies that .
The construction of proves that its image is contained in the commutative subalgebra B, which is contained in the bicommutant of I, p. 6).
If the radical of the algebra A is zero, the uniqueness of the morphism is valid without requiring that it be continuous (cf. prop. 9 of I, p. 40). This is not so in general, cf. G. R. Allan, Embedding the algebra of formal power series in a Banach algebra, Proc. London Math. Soc. (3) 25 (1972), 329–340.
For every Banach algebra A, every element of A and every germ (Sp), one denotes by the element of theorem 5. If A is a commutative Banach algebra, this element coincides with the element furnished by the holomorphic functional calculus on a commutative Banach algebra (theorem 1 of I, p. 51).
Let B be the closed full subalgebra of A generated by , so that Sp Sp. The element of A belongs to B, and coincides with the element calculated relative to the algebra B.
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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