By Browder F. (ed.)
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Additional resources for Nonlinear Functional Analysis and Its Applications, Part 1
Example text
Proof. Let A ∈ I. 54 yields Sup A = A. It follows −A = − Sup A = Inf(−A) and hence −A ∈ I. Note that the last statement is not true for A ∈ F . Nevertheless, it is sometimes easier to work with the complete lattice F in the proofs. 52. In the following proposition we use a generalization of the Minkowski sum. For A, B ⊆ I, we set A ⊕ B := {I ∈ I| ∃A ∈ A, ∃B ∈ B : I = A ⊕ B} . 56. Let A, B ⊆ I, then (i) (ii) inf A ⊕ B = inf A ⊕ inf B, sup A ⊕ B sup A ⊕ sup B . Proof. (i) If A = ∅, we have inf A ⊕ B = inf A = {+∞} and thus inf A ⊕ B = inf A ⊕ inf B = {+∞}.
Hence we have {f (¯ ¯ for (ii) ⇒ (i). By (ii), f is constant on X. x)} = f [X] ¯ all x¯ ∈ X. 1). 13. 1). 1) for every x ¯ ∈ X. 1) if and only if {¯ x} is a unique solution to (L). Proof. (i) is equivalent to ¯ ⊆S X ∧ inf f (x) = inf f (x) ¯ x∈X x∈S ∧ ¯ = Min f [S]. 12, this is equivalent to ¯: ∀¯ x∈X x ¯∈S ∧ f (¯ x) = inf f (x) x∈S ∧ {f (¯ x)} = Min f [S], which is an alternative way to express (ii). 2 (where a complete lattice Z is obtained by extending R2 by two elements ±∞), Eff (L) is not a solution to (L); whence a solution does not exist.
Let Y be a partially ordered topological vector space with an ordering cone C such that ∅ = int C = Y and let F = FC (Y ), then Fco = {A ⊆ Y | Cl + co A = A} . Proof. We have A = Cl + co A ⇐⇒ A = co A ∧ A = Cl + A ⇐⇒ ∀λ ∈ [0, 1] : A = Cl + λA + (1 − λ)A ⇐⇒ ∀λ ∈ [0, 1] : A = λ A ⊕ (1 − λ) A. 15. 30), we can also work with the convex hull of a subset B ⊆ Y and we obtain a similar characterization of the conlinear space Ico . To this end we need the following result. 60. Let Y be an extended partially ordered topological vector space with an ordering cone C such that ∅ = int C = Y .