Let G be a Lie group, E1,…,En+1 complete normable spaces and π1,…,πn+1 analytic linear representations of G on E1,…,En+1. Let
E=L(E1,…,En;En+1) the complete normable space of continuous multilinear mappings of E1×⋯×En into En+1 (General Topology, Chapter X, § 3, no. 2).
For all g∈G,
let π(g) be the automorphism of E defined by (π(g)u)(x1,…,xn)=πn+1(g)(u(π1(g)−1x1,…,πn(g)−1xn)).
Then π is an analytic linear representation of G on E and
((L(π)a)u)(x1,…,xn)=−i=1∑nu(x1,…,xi−1,(L(πi)a)xi,xi+1,…,xn) +(L(πn+1)a)(u(x1,…,xn)) for all
a∈L(G),
u∈E,
x1∈E1,…,xn∈En.
Every element (A1,…,An+1) of L(E1)×⋯×L(En+1) defines a continuous endomorphism θ(A1,…,An+1) of E by the formula
(θ(A1,…,An+1)u)(x1,…,xn)=An+1(u(A1x1,…,Anxn)). The mapping
θ of
L(E1)×⋯×L(En+1) into
L(E) is continuous and multilinear. Then, for all
g∈G,
π(g)=θ(π1(g−1),…,πn(g−1),πn+1(g)) and hence
π is analytic. We apply Proposition 41 to the mapping
(x1,…,xn,u)↦u(x1,…,xn) of
E1×⋯×En×E into
En+1. Then
πn+1(g)(u(x1,…,xn))=(π(g)u)(π1(g)x1,…,πn(g)xn) and hence
(L(πn+1)a)(u(x1,…,xn))=i=1∑nu(x1,…,(L(πi)a)xi,…,xn) +((L(π)a)u)(x1,…,xn).
When the Ei are finite-dimensional, the representation L(π) of L(G) is derived from the representations L(π1),…,L(πn+1) under the procedure of Chapter I, § 3, Proposition 3.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.