Bourbaki

Proposition 1

Tag 02A5, permanent. Cite it as [Bourbaki, Tag 02A5].

Théories spectrales, chapter I, § 4. Calcul fonctionnel holomorphe

Let aa be an element of A, let U be an open neighbourhood of Sp(aa), and let f∈O(U;A)f\in \mathscr{O}(U; A). Let V be a compact neighbourhood of Sp(aa) contained in U and such that V is a piece of C\mathbf{C}.

Then z↦f(z)(z−a)−1z\mapsto f(z)(z-a)^{-1} is continuous on ∂V\partial V, the differential form f(z)(z−a)−1dzf(z)(z-a)^{-1}dz is integrable on ∂V\partial V, and one has

Θa(f)=12iπ∫∂Vf(z)(z−a)−1dz \Theta_a(f) =\frac{1}{2i\pi}\int_{\partial V}f(z)(z-a)^{-1}dz

Let hh be a mapping of C\mathbf{C} into C\mathbf{C}, indefinitely differentiable, equal to 1 in a neighbourhood of Sp(aa) and with compact support contained in the interior of V (lemma 1 of I, p. 52). Let uu be a mapping of C\mathbf{C} into A such that (h,u)(h, u) is adapted to a(a(cf. example 3 of I, p. 53). The associated differential form is ω=du∧dz\omega =du\wedge dz. We get fω=fdu∧dz=d(fudz)f \omega =f du\wedge dz=d(f u dz) since ff is holomorphic. Moreover, u(z)=(z−a)−1u(z) = (z-a)^{-1} on the boundary of V. On the other hand, the differential form fudzf u dz is of class C1C^1 on U. Hence

2iπΘa(f)=∫Vd(fudz)=∫∂Vfudz=∫∂Vf(z)(z−a)−1dz 2i\pi \Theta_a(f) =\int_Vd(f u dz) =\int_{\partial V}f u dz=\int_{\partial V}f(z)(z-a)^{-1}dz

by formula (5) and Stokes's formula for the piece V (VAR, R2, p. 47, 11.2.3).

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.

Cited by

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Cites

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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.