Lemma 2
Tag 03AA, permanent. Cite it as [Bourbaki, Tag 03AA].
Théories spectrales, chapter V, § 2. Représentations des groupes localement compacts
Let be a nonzero unitary representation of G isomorphic to a subrepresentation of . There exists a measure such that is a nonzero hermitian compact endomorphism of the space of . In particular, there exists a nonzero real number such that the eigensubspace of corresponding to is nonzero.
One may suppose that is a subrepresentation of . Let be its space. For every measure , the endomorphism is deduced from by passing to subspaces. The hypothesis and prop. 3 of III, p. 5 therefore imply that is compact.
Since converges to the measure in endowed with the topology of compact convergence and since the space is nonzero, there exists such that is a nonzero endomorphism of (cf. No. of V, p. 400). The endomorphism is equal to (lemma 1 of V, p. 401); it is therefore nonzero (EVT, V, p. 39, prop. 2), hermitian and compact, since is compact.
Since is hermitian and nonzero, its spectrum is not reduced to zero (example 1 of I, p. 110). Finally, since is compact, every Sp() is an eigenvalue of (prop. 2 of III, p. 83).
Let us now prove the proposition.
We shall use prop. 7 of V, p. 391 to establish that is semisimple. Let be a nonzero subrepresentation of , and its space; we must prove that contains an irreducible subrepresentation.
Let be such that is a nonzero compact hermitian endomorphism of the space of , and let be nonzero such that the eigensubspace F of relative to is nonzero (lemma 2). The space F is of finite dimension (prop. 5 of III, p. 90). There exists a subrepresentation of on a subspace of such that is nonzero and of minimal dimension. Let be a nonzero element of , and let be the subrepresentation of generated by . Then , whence by minimality of the dimension of .
Let us prove that the representation is irreducible. Let be a closed subspace stable under G. We have , where denotes the orthogonal of in (in fact, if , let be its orthogonal projection on ; since and are stable under , the vector is the orthogonal projection of on , whence , which means that , and the assertion follows). The minimality of the dimension of then implies that either , or . In the first case, we have , whence , and in the second, we have , whence and . The representation is therefore irreducible.
It follows from Prop. 7 of V, p. 391 that the representation is semisimple.
Let be an irreducible unitary representation of G whose multiplicity in is nonzero; let be its space. There exist a measure and a nonzero real number such that is a nonzero compact hermitian endomorphism of and that the proper subspace of corresponding to is nonzero (Lemma 2). Let F be the proper subspace of corresponding to . For every subrepresentation of E isomorphic to , the endomorphism of deduced from by passing to the subspaces is identified with , hence the intersection of and F is nonzero. Thus, the multiplicity of in is less than the dimension of the space F, which is finite (Prop. 5 of III, p. 90).
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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