Lecture 3

In particular c a z for all z c since c is a cone it

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Unformatted text preview: co C )∗ , and (iii) C ∗ ⊃ D∗ . Proof. C ∗ is nonempty since 0 ∈ C . Since the half space − Hx := y ∈ RN ⟨x, y ⟩ ≤ 0 − is a closed convex cone (easy), so is C ∗ = x∈C Hx . If y ∈ D∗ , then ⟨x, y ⟩ ≤ 0 for all x ∈ D. Since C ⊂ D, we have ⟨x, y ⟩ ≤ 0 for all x ∈ C . Therefore y ∈ C ∗ , which proves D∗ ⊂ C ∗ . Finally, since C ⊂ co C , we have C ∗ ⊃ (co C )∗ . To prove the reverse inclusion, take any x ∈ co C . By Lemma 1, there exists a K convex combination x = k=1 αk xk such that xk ∈ C for all k . If y ∈ C ∗ , it follows that ⟨x, y ⟩ = αk xk , y = αk ⟨xk , y ⟩ ≤ 0, so y ∈ (co C )∗ . Therefore C ∗ ⊂ (co C )∗ . Proposition 6. Let C ⊂ RN be a nonempty cone. Then C ∗∗ = cl co C . (C ∗∗ is the dual cone of C ∗ , so it is the dual cone of the dual cone of C .) Proof. Let x ∈ co C . For any y ∈ C ∗ = (co C )∗ we have ⟨x, y ⟩ ≤ 0. This implies x ∈ C ∗∗ . Hence co C ⊂ C ∗∗ . Since the dual cone is closed, we have cl co C ⊂ cl C ∗∗ = C ∗∗ . To show the reverse inclusion, suppose that x ∈ cl co C . Then by Proposition / 4 there exists a separating hyperplane such that ⟨a, x⟩ > c > ⟨a, z ⟩ 2014W Econ 172B Operations Research (B) Alexis Akira Toda for any z ∈ co C . In particular, c > ⟨a, z ⟩ for all z ∈ C . Since C is a cone, it must be ⟨a, z ⟩ ≤ 0 for all z , for otherwise if there is z0 ∈ C with ⟨a, z0 ⟩ > 0, for any β > 0 we have β z0 ∈ C , but for large enough β we have ⟨a, β z0 ⟩ = β ⟨a, z0 ⟩ > c, a contradiction. This shows that a ∈ C ∗ . Again since C is a cone, we have 0 ∈ C , so c > ⟨a, 0⟩ = 0. Since ⟨a, x⟩ > c >= and a ∈ C ∗ , it follows that x ∈ C ∗∗ . Therefore C ∗∗ ⊂ cl co C . / The following corollary will play an important role in optimization theory. Corollary 7 (Farkas). Let C = cone[a1 , . . . , aK ] be a polyhedral cone generated by a1 , . . . , aK . Let D = y ∈ RN (∀k ) ⟨ak , y ⟩ ≤ 0 . Then D = C ∗ and C = D∗ . Proof. Let y ∈ D. For any x ∈ C , we can take {αk } ≥ 0 such that x = Then ⟨x, y ⟩ = αk ⟨ak , y ⟩ ≤ 0, k αk ak . k so y ∈ C . Conversely, let y ∈ C . Since ak ∈ C , we get ⟨ak , y ⟩ ≤ 0 for all k , so y ∈ D. Therefore D = C ∗ . Since C is a closed convex cone, by Propositions 6 and 5 (iii) we get ∗ ∗ C = cl co C = C ∗∗ = (C ∗ )∗ = D∗ . 4 4.1 Applications to economics and finance Pareto efficient allocations Suppose that there are I consumers indexed by i = 1, . . . , I with continuous utility functions ui : RN...
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