Bourbaki

Theorem 2

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

Théories spectrales, chapter IV, § 1. Opérateurs compacts sur un espace hilbertien

Suppose that uNu_N is of finite trace. Then the function x↦N(x,x)x\mapsto N(x, x) belongs to L1(X)\mathscr{L}^1(X) and we have

Tr(uNu_N) =∫XN(x,x)dμ(x)=\int_XN(x, x)d\mu(x).

From conditions (i) and (ii), we have ∣N(x,x)∣⩽H(x,x)|N(x, x)|\leqslant H(x, x) for x∈X~x\in \widetilde{X}, therefore the function x↦N(x,x)x\mapsto N(x, x) belongs to L1(X)\mathscr{L}^1(X) by condition (iii).

Condition (i) establishes the equality

N(x,x)=∑i∈Iαifi(x)‾gi(x) N(x, x) =\sum_{i\in I}\alpha_i\overline{f_i(x)}g_i(x)

for all x∈X~x\in \widetilde{X}. From conditions (ii) and (iii), we may apply Lebesgue's theorem (INT, IV, p. 137, § 3, no7^o7, th. 6), from which it follows that

∫XN(x,x)dμ(x)=∑i∈Iαi∫Xfi(x)‾gi(x)dμ(x) \int_XN(x, x)d\mu(x) =\sum_{i\in I}\alpha_i\int_X\overline{f_i(x)}g_i(x)d\mu(x)

=∑i∈Iαi⟨fi∣gi⟩==\sum_{i\in I}\alpha_i\langle f_i|g_i\rangle = Tr(uNu_N),

the last equality resulting from the cor. of prop. 17 of IV, p. 170.

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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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.