Lemma 4

Tag 02U0, permanent. Cite it as [Bourbaki, Tag 02U0].

Théories spectrales, chapter III, § 5. Perturbation par une application linéaire compacte

The linear mapping from $G_{\lambda}$ into $E'$ defined by $g\mapsto g\cdot \mu$ is injective, and its image is the kernel of $1_{E'}-\lambda^th$ .

If $g\in G_{\lambda}$ satisfies $g\cdot \mu= 0$ , then formula (5) implies $g= 0$ , hence the mapping $g\mapsto g\cdot \mu$ is injective.

Let $m\in E'$ be a measure belonging to the kernel of $1_{E'}-\lambda^th$ . According to relation (6), one has

\int_Xf(x)dm(x) =\lambda \int_X(\int_XN(x, y)dm(x))f(y)d\mu(y) \tag{7}

for every continuous function $f: X\rightarrow K$ . The measure $m$ is therefore equal to the measure $g\cdot \mu$ , where $g$ is the continuous function from X into K defined by

g(y) =\lambda \int_XN(x, y)dm(x) \tag{8}

for every $y\in X$ . Formula (8) then implies that the function $g$ belongs to $G_{\lambda}$ .

Conversely, let $g: X\rightarrow K$ be a continuous function belonging to $G_{\lambda}$ and put $m=g\cdot \mu$ . For every continuous function $f: X\rightarrow K$ , we therefore have

\int_Xf(y)dm(y) =\int_Xg(y)f(y)d\mu(y)

=\lambda \int_X(\int_XN(x, y)g(x)d\mu(x))f(y)d\mu(y)

=\lambda \int_Xf(x)dm(x)

so that $^th(m) =\lambda m$ . This proves Lemma 4.

By applying Proposition 4 and Theorem 4 of III, p. 78. to the endomorphism $h$ , one then obtains the following statements.

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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