Assume that C is weakly complete. Let A be the set of restrictions to C of the continuous linear forms on E. Let (fλ)λ∈Λ be a finite family of elements of A, and f=sup(fλ). For every x∈C, define
f(x)=sup(f(x1)+f(x2)+⋯+f(xn)),
*the supremum being taken over the set Sx of sequences (x1,x2,…,xn) of elements of C such that x1+x2+⋯+xn=x. Let CardΛ=p. Then there exists (y1,…,yp)∈Sx such that f(x)=f(y1)+⋯+f(yp).
Denote by f1,f2,…,fp the elements of the family (fλ). For k=1,2,…,p, let Ck be the set of y∈C such that
f1(y)<f(y), f2(y)<f(y), …, fk−1(y)<f(y), fk(y)=f(y).
The Ck are disjoint convex cones with union C. Let x1,x2,…,xn in C be such that x1+x2+⋯+xn=x. Let yk be the sum of the xi that belong to Ck. Then y1+y2+⋯+yp=x. Since f is affine on Ck, f(y1)+⋯+f(yp)=f(x1)+⋯+f(xn). Therefore
f(x)=sup(f(y1)+⋯+f(yp)),
where (y1,y2,…,yp) runs over the set of sequences of p points of C such that y1+y2+⋯+yp=x. Set D=C∩(x−C). Since D is compact (TVS, II, §6, No. 8, Cor. 2 of Prop. 11), so is the set of elements (y1,…,yp) of Dp such that y1+⋯+yp=x, thus the supremum (19) is attained.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.