Anderson's lemma

Appendix DVector LatticesProof.Define the u ij as in the following table:u 11 = x 1 ∧ y 1 u 12 = (y 1 − x 1 ) + y 1u 21 = (y 2 − x 2 ) + u 22 = y 2 ∧ x 2 y 2x 1 x 2 x 1 + x 2 = y 1 + y 2Check the first row sum:x 1 ∧ y 1 + (y 1 − x 1 ) + = (x 1 ∧ y 1 ) + (y 1 ∨ x 1 ) − x 1 = x 1 + y 1 − x 1Interchange subscripts 1 and 2 to check the second row sum. Interchange theroles of x and y, then use the fact that y 2 − x 2 = x 1 − y 1 to check the column□ sums.splitting.ineq Corollary. If w, x 1 , x 2 ∈ V + and w ≤ x 1 + x 2 then there exist w i ∈ V +such that w = w 1 + w 2 and w i ≤ x i for i = 1, 2.□ Proof. let y 1 = w and y 2 = x 1 + x 2 − w.sum.orthog Exercise. Suppose x ∈ V and y 1 , y 2 ∈ V + and x ⊥ y i for i = 1, 2. Showthat x ⊥ (y 1 + y 2 ) and x ⊥ (y 1 − y 2 ).Solution: Suppose 0 ≤ w =|x|∧(y 1 + y 2 ). Split w into a sum w 1 + w 2 ,with 0 ≤ w i ≤ y i . Note w i ≤|x|.w = w 1 ∧|x|+w 2 ∧|x| ≤y 1 ∧|x|+y 2 ∧|x| =0.Thus |x| ⊥(y 1 + y 2 ).The second assertion follows from the first and the fact that |y 1 − y 2 |≤□ y 1 + y 2 .[§] 3. Order-bounded linear functionalslet V be a vector lattice. A linear map λ from V into R is to be an orderbounded linear functional if, for each pair a ≤ b,sup{|λ(x)| : a ≤ x ≤ b} < ∞A linear map λ from V into R is to be an increasing linear functional ifλ(x) ≤ λ(y) whenever x ≤ y. Equivalently, a linear functional is increasing ifλ(x) ≥ 0 for all x ∈ V + . The space V # of all order-bounded linear functionalson V is called the order dual of V .fnal.ordering Definition. Define λ 1 ≤ λ 2 to mean that λ 1 (x) ≤ λ 2 (x) for all x in V + .The key facts about linear functionals are that(i) The order dual V # is a vector lattice.(ii) Each λ in V # can be expressed as a difference of two increasing linearfunctionals.extension Lemma. Let λ : V + → R satisfy(i) λ(αx + βy) = αλ(x) + βλ(y) for all α, β ∈ R + and x, y ∈ V +(ii) For each w in V + ,sup{|λ(x)| :0≤ x ≤ w} < ∞Then λ has a unique extension to an order-bounded linear functional on V .Proof. Define λ(u) = λ(u + ) − λ(u − ). Check that if x 1 − x 2 = y 1 − y 2 thenλ(x 1 ) − λ(x 2 ) = λ(y) 1 − λ(y 2 ). Deduce linearity.Asymptopia: 4Jan99 c○David Pollard 3