Bourbaki

Lemma 5

Tag 08ZA, permanent. Cite it as [Bourbaki, Tag 08ZA].

Integration, chapter IV, § 7. Barycenters

Assume that C is weakly complete. Let A\mathscr{A} be the set of restrictions to C of the continuous linear forms on E. Let (fλ)λ∈Λ(f_\lambda)_{\lambda \in \Lambda} be a finite family of elements of A\mathscr{A}, and f=sup⁡(fλ)f = \sup(f_\lambda). For every x∈Cx \in C, define

f‾(x)=sup⁡(f(x1)+f(x2)+⋯+f(xn)), \overline{f}(x) = \sup \left( f(x_1) + f(x_2) + \cdots + f(x_n) \right),

*the supremum being taken over the set SxS_x of sequences (x1,x2,…,xn)(x_1, x_2, \ldots, x_n) of elements of C such that x1+x2+⋯+xn=xx_1 + x_2 + \cdots + x_n = x. Let Card Λ=p\mathrm{Card}\, \Lambda = p. Then there exists (y1,…,yp)∈Sx(y_1, \ldots, y_p) \in S_x such that f‾(x)=f(y1)+⋯+f(yp)\overline{f}(x) = f(y_1) + \cdots + f(y_p).

Denote by f1,f2,…,fpf_1, f_2, \ldots, f_p the elements of the family (fλ)(f_\lambda). For k=1,2,…,pk = 1, 2, \ldots, p, let CkC_k be the set of y∈Cy \in C such that

f1(y)<f(y), f2(y)<f(y), …, fk−1(y)<f(y), fk(y)=f(y). f_1(y) < f(y),\ f_2(y) < f(y),\ \ldots,\ f_{k-1}(y) < f(y),\ f_k(y) = f(y).

The CkC_k are disjoint convex cones with union C. Let x1,x2,…,xnx_1, x_2, \ldots, x_n in C be such that x1+x2+⋯+xn=xx_1 + x_2 + \cdots + x_n = x. Let yky_k be the sum of the xix_i that belong to CkC_k. Then y1+y2+⋯+yp=xy_1 + y_2 + \cdots + y_p = x. Since ff is affine on CkC_k, f(y1)+⋯+f(yp)=f(x1)+⋯+f(xn)f(y_1) + \cdots + f(y_p) = f(x_1) + \cdots + f(x_n). Therefore

f(x)=sup⁡(f(y1)+⋯+f(yp)), f(x) = \sup \left( f(y_1) + \cdots + f(y_p) \right),

where (y1,y2,…,yp)(y_1, y_2, \ldots, y_p) runs over the set of sequences of pp points of C such that y1+y2+⋯+yp=xy_1 + y_2 + \cdots + y_p = x. Set D=C∩(x−C)D = C \cap (x - C). Since D is compact (TVS, II, §6, No. 8, Cor. 2 of Prop. 11), so is the set of elements (y1,…,yp)(y_1, \ldots, y_p) of DpD^p such that y1+⋯+yp=xy_1 + \cdots + y_p = 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.

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Transcribed from the printed volume. Not translated.