Proposition 22

Tag 02Y0, permanent. Cite it as [Bourbaki, Tag 02Y0].

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

Suppose that $u_N$ is a positive endomorphism of $L^2(X)$ . Then the endomorphism $u_N$ is of finite trace if and only if the function $x\mapsto N(x, x)$ is $\mu$ -integrable. In that case, one has

Tr( $u_N$ ) $=\int_XN(x, x)d\mu(x)$ .

If $u_N$ is of finite trace, Theorem 2 implies that $x\mapsto N(x, x)$ is integrable on X and that its integral is the trace of $u_N$ .

Conversely, suppose that the function $x\mapsto N(x, x)$ is integrable. Since $u_N$ is positive, the orthonormal families $(f_i)_{i\in I}$ and $(g_i)_{i\in I}$ can be chosen so that $f_i=g_i$ for all $i\in I$ (remark 3 of IV, p. 151). For every $x\in \widetilde{X}$ , one has

N(x, x) =\sum_{i\in I}\alpha_i|f_i(x)|^2

whence, for every finite subset J of I, one deduces

\sum_{i\in J}\alpha_i=\sum_{i\in J}\alpha_i\int_X|f_i(x)|^2d\mu(x)

=\int_X\sum_{i\in J}\alpha_i|f_i(x)|^2d\mu(x)\leqslant \int_XN(x, x)d\mu(x)

The family $(\alpha_i)_{i\in I}$ is therefore summable, which implies that the endomorphism $u_N$ is of finite trace (prop. 12 of IV, p. 164).

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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