Bourbaki

Exercise 6, § 2

Integration, II, p. 20

Bourbaki marks this one as harder than the rest.

Let E be a Riesz space, Ω\Omega the fully lattice-ordered space of relatively bounded linear forms on E, and F a co-lattice subspace of Ω\Omega (§1§ 1, Exer. 3). For every x∈Ex \in E, the mapping x′↦⟨x,x′⟩x' \mapsto \langle x, x' \rangle of F into R\mathbf{R} is a relatively bounded linear form uxu_x on F, and the mapping x↦uxx \mapsto u_x is an increasing linear mapping of E into the fully lattice-ordered space Ω′\Omega' of relatively bounded linear forms on F. In order that x↦uxx \mapsto u_x be an isomorphism of E onto a co-lattice subspace of Ω′\Omega', that is, for ux>0u_x > 0 to imply x>0x > 0, and that usup⁡(x,y)=sup⁡(ux,uy)u_{\sup(x, y)} = \sup(u_x, u_y), it is necessary and sufficient that the following two conditions be verified: 1° for every x>0x > 0 in E, there exists an x′>0x' > 0 in F such that ⟨x,x′⟩>0\langle x, x' \rangle > 0; 2° for every pair of alien elements y≥0,z≥0y \geq 0, z \geq 0 of E, for every number ε>0\varepsilon > 0 and for every x′≥0x' \geq 0 in F, there exist two elements y′≥0,z′≥0y' \geq 0, z' \geq 0 of F such that x′=y′+z′x' = y' + z' and ⟨y,y′⟩+⟨z,z′⟩≤ε\langle y, y' \rangle + \langle z, z' \rangle \leq \varepsilon. (Observe that if the second condition is verified and if vv is a positive linear form on F such that v≥uxv \geq u_x, then v(x′)≥⟨x+,x′⟩−εv(x') \geq \langle x^+, x' \rangle - \varepsilon for every x′≥0x' \geq 0 in F and every ε>0\varepsilon > 0.)

If the condition 2° is fulfilled but not condition 1°, then the set of x∈Ex \in E such that ⟨x,x′⟩=0\langle x, x' \rangle = 0 for all x′∈Fx' \in F is an isolated linear subspace H of E. By passage to the quotient, the mapping x↦uxx \mapsto u_x then defines an isomorphism of the Riesz space E/H onto a co-lattice subspace of Ω′\Omega'.

Show that when F=ΩF = \Omega, the above condition 2° is always verified (make use of Exer. 2).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.