Bourbaki

Exercise 4, § 2

Integration, II, p. 19

Bourbaki marks this one as harder than the rest.

Let E be a fully lattice-ordered space.

a) Let LL be a positive linear form on E, continuous for the topology T0(E)\mathscr{T}_0(E) (§1, Exer. 10). Show that the set of x∈Ex \in E such that L(∣x∣)=0L(|x|) = 0 is a band Z(L)Z(L). Let S(L)S(L) be the band supplementary to Z(L)Z(L) in E (§1, Exer. 6).

b) Let LL and MM be two positive linear forms on E, continuous for T0(E)\mathscr{T}_0(E). In order that, in the space Ω\Omega of relatively bounded linear forms on E, MM belong to the band generated by LL, it is necessary and sufficient that S(M)⊂S(L)S(M) \subset S(L). (To see that the condition is sufficient, argue by contradiction, on supposing that MM does not verify the condition of Prop. 5; consider a sequence (yn)(y_n) of elements of E such that 0≤yn≤x0 \leq y_n \leq x,

L(yn)≤1/2nL(y_n) \leq 1/2^n and M(yn)≥α>0M(y_n) \geq \alpha > 0, and infer the existence of an element z≥0z \geq 0 in E such that L(z)=0L(z) = 0 and M(z)≥α.M(z) \geq \alpha .

c) Let LL and MM be two positive linear forms on E, continuous for T0(E)\mathscr{T}_0(E). In order that, in Ω\Omega, LL and MM be alien, it is necessary and sufficient that S(L)∩S(M)={0}S(L) \cap S(M) = \{0\}. (To see that the condition is necessary, prove that if S(L)∩S(M)S(L) \cap S(M) does not reduce to 0 then the bands in Ω\Omega generated by LL and MM have an element ≠0\neq 0 in common, by using b.)

Solution

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