Bourbaki

Lemma 1

Tag 035C, permanent. Cite it as [Bourbaki, Tag 035C].

Théories spectrales, chapter IV, § 5. Opérateurs partiels normaux et théorème spectral

The subset Ω(E) is self-adjoint in L(E)\mathscr{L}(E).

Let v∈Ω(E)v\in \Omega (E). One has ∥v∗∥=∥v∥⩽1\|v^*\|=\|v\|\leqslant 1 therefore the endomorphism 1E−vv∗1_E-vv^* is positive. It is injective: if x∈x\in Ker(1E−vv∗_E-vv^*), one has vv∗(x)=xvv^*(x) =x, whence v∗(v(v∗(x)))=v∗(x)v^*(v(v^*(x))) =v^*(x) then v∗(x)=0v^*(x) = 0, since 1E−v∗v1_E-v^*v is injective, and finally x=v(v∗(x))=0x=v(v^*(x)) = 0. The lemma follows.

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.

Cited by

Nothing.

Cites

Nothing.

Machine translation of the English, by gpt-5.4, gpt-5-6-mini, 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.