Bourbaki

Proposition 9

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

Théories spectrales, chapter III, § 3. Endomorphismes de Fredholm et endomorphismes de Riesz

Let E be a separated locally convex space and let u,vu,v be permutable elements of L(E)\mathscr{L}(E). The following conditions are equivalent:

(i) The endomorphisms uu and vv are Riesz endomorphisms;

(ii) The endomorphism u∘vu\circ v is a Riesz endomorphism.

Suppose that uu and vv are Riesz endomorphisms; let u′u' and v′v' denote their canonical quasi-inverses. The endomorphisms u,v,u′u,v,u' and v′v' of E commute (prop. 6, d)). The endomorphism v′∘u′v'\circ u' is a quasi-inverse of u∘vu\circ v which commutes with u∘vu\circ v, hence u∘vu\circ v is a Riesz endomorphism (prop. 8).

Suppose conversely that u∘vu\circ v is a Riesz endomorphism, and let ww be its canonical quasi-inverse. Since uu commutes with u∘vu\circ v, the endomorphisms uu and ww commute by prop. 6, d). Analogously vv and ww commute. Hence the endomorphisms u,vu,v and ww of E commute; it follows that the endomorphism v∘wv\circ w of E is a quasi-inverse of uu and that w∘uw\circ u is a quasi-inverse of vv. By prop. 8,u,u and vv are Riesz endomorphisms.

Let uu be a Riesz endomorphism of E. If vv is an automorphism of E, the endomorphism u∘vu\circ v of E is not necessarily a Riesz endomorphism of E, even if u∘v−v∘uu\circ v-v\circ u is of finite rank. If hh is an endomorphism of finite rank of E, the endomorphism u+hu+h is not necessarily a Riesz endomorphism (cf. III, p. 120, exercise 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.

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Machine translation of the English, by gpt-5-6-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.