Bourbaki

Proposition 7

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

Théories spectrales, chapter III, § 4. Perturbations dans les espaces de Banach

Let E be a Banach space and F a normed space. The mapping u↦((u))u\mapsto ((u)) from L(E;F)\mathscr{L}(E; F) into R‾\overline{\mathbf{R}} is upper semi-continuous.

Let u∈L(E;F)u\in \mathscr{L}(E; F). We have to prove that for every real number c>((u))c >((u)), the set of elements v∈L(E;F)v\in \mathscr{L}(E; F) such that ((v))<c((v))< c is a neighbourhood of uu. Let B denote the set of elements of E of norm <1<1. By Prop. 6, there exists y∈Ey\in E such that u(y)∉u(B)u(y)\notin u(B) and ∥u(y)∥<c\|u(y)\|< c. The distance dd from u(y)u(y) to the closed set u(B)u(B) is strictly positive. The set V of elements vv of L(E;F)\mathscr{L}(E; F) satisfying the relations ∥v(y)∥<c\|v(y)\|< c and ∥u−v∥(1+∥y∥)<d\|u-v\|(1 +\|y\|)< d is a neighbourhood of uu in L(E;F)\mathscr{L}(E; F). Let v∈Vv\in V. For every x∈Bx\in B, one has

d⩽∥u(y)−u(x)∥⩽∥v(y)−v(x)∥+∥u−v∥(∥y∥+∥x∥) d\leqslant \|u(y)-u(x)\|\leqslant \|v(y)-v(x)\|+\|u-v\|(\|y\|+\|x\|)
<∥v(y)−v(x)∥+d <\|v(y)-v(x)\|+d

Consequently v(y)v(y) does not belong to vv(B), and one has ((v))⩽∥v(y)∥<c((v))\leqslant \|v(y)\|< c by Prop. 6.

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, against glossary version 34. Not checked by a person.