v2010.10.26 - Convex Optimization

2.13. DUAL CONE & GENERALIZED INEQUALITY 177xKFigure 61: x ≽ 0 with respect to K but not with respect to nonnegativeorthant R 2 + (pointed convex cone K drawn truncated).2.13.4.2.2 Exercise. Discretized dual generalized inequalities.Test Theorem 2.13.4.2.1 on Figure 56a using extreme directions there asgenerators.From the discretized membership theorem we may further deduce a moresurgical description of closed convex cone that prescribes only a finite numberof halfspaces for its construction when polyhedral: (Figure 55a)K = {x ∈ R n | 〈γ ∗ , x〉 ≥ 0 for all γ ∗ ∈ G(K ∗ )} (368)K ∗ = {y ∈ R n | 〈γ , y〉 ≥ 0 for all γ ∈ G(K)} (369)2.13.4.2.3 Exercise. Partial order induced by orthant.When comparison is with respect to the nonnegative orthant K = R n + , thenfrom the discretized membership theorem it directly follows:x ≼ z ⇔ x i ≤ z i ∀i (370)Generate simple counterexamples demonstrating that this equivalence withentrywise inequality holds only when the underlying cone inducing partialorder is the nonnegative orthant; e.g., explain Figure 61.