Let G be a complete normed algebra over R, and a(t), b(t) and c(t) three maps from J into G; the equation
dtdx=a(t)x+xb(t)+c(t)
is a linear differential equation; here A(t) is the linear map x↦a(t)x+xb(t) of G to itself.
For every t∈J, A(t) is an element of the set L(E) of continuous linear maps from E to itself (continuous endomorphisms of E); one knows (Gen. Top., X, p. 298) that L(E), endowed with the norm ∥U∥=sup∥x∥≤1∥Ux∥ is a complete normed algebra over the field R, and that ∥UV∥≤∥U∥ ∥V∥.
Throughout this section we shall assume that the following conditions are satisfied:
a) The map t↦A(t) of J into L(E) is regulated.
b) The map t↦b(t) of J into E is regulated.
When E has dimension n, L(E) is isomorphic to Rn2 (as a topological vector space) and condition a) means that each of the elements aij(t) of the matrix A(t) is a regulated function on J.
Since ∥A(t′)x−A(t)x∥≤∥A(t′)−A(t)∥∥x∥, the map
t↦A(t).x+b(t) is regulated for every
x∈E; further,
∥A(t)x1−A(t)x2∥=∥A(t)(x1−x2)∥≤∥A(t)∥∥x1−x2∥ for any
t∈J and
x1,x2 in
E; in other words, the right-hand side of (2) satisfies the conditions of lemma 1 of IV, p. 165 and is Lipschitz with respect to the regulated function
∥A(t)∥ on
J×E. In consequence (IV, p. 173, cor. 2):
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.