Let E be a Banach space and F a normed space. The mapping u↦((u)) from L(E;F) into R is upper semi-continuous.
Let u∈L(E;F). We have to prove that for every real number c>((u)), the set of elements v∈L(E;F) such that ((v))<c is a neighbourhood of u. Let B denote the set of elements of E of norm <1. By Prop. 6, there exists y∈E such that u(y)∈/u(B) and ∥u(y)∥<c. The distance d from u(y) to the closed set u(B) is strictly positive. The set V of elements v of L(E;F) satisfying the relations ∥v(y)∥<c and ∥u−v∥(1+∥y∥)<d is a neighbourhood of u in L(E;F). Let v∈V. For every x∈B, one has
d⩽∥u(y)−u(x)∥⩽∥v(y)−v(x)∥+∥u−v∥(∥y∥+∥x∥)
<∥v(y)−v(x)∥+d
Consequently v(y) does not belong to v(B), and one has ((v))⩽∥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.