v2007.09.13 - Convex Optimization

v2007.09.13 - Convex Optimization v2007.09.13 - Convex Optimization

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140 CHAPTER 2. CONVEX GEOMETRYthe primal program. Third, the maximum value achieved by the dual problemis often equal to the minimum of the primal. [221,2.1.3] Essentially, dualitytheory concerns representation of a given optimization problem as half aminimax problem. [228,36] [46,5.4] Given any real function f(x,z)minimizexalways holds. Whenminimizexmaximizezmaximizezf(x,z) ≥ maximizezf(x,z) = maximizezminimize f(x,z) (261)xminimize f(x,z) (262)xwe have strong duality and then a saddle value [103] exists. (Figure 45)[225, p.3] Consider primal conic problem (p) and its corresponding dualproblem (d): [216,3.3.1] [174,2.1] given vectors α , β and matrixconstant C(p)minimize α T xxsubject to x ∈ KCx = βmaximize β T zy , zsubject to y ∈ K ∗C T z + y = α(d) (263)Observe the dual problem is also conic, and its objective function value neverexceeds that of the primal;α T x ≥ β T zx T (C T z + y) ≥ (Cx) T zx T y ≥ 0(264)which holds by definition (258). Under the sufficient condition: (263p) isa convex problem and satisfies Slater’s condition, 2.46 then each problem (p)and (d) attains the same optimal value of its objective and each problemis called a strong dual to the other because the duality gap (primal−dualobjective difference) is 0. Then (p) and (d) are together equivalent to theminimax problemminimize α T x − β T zx,y,zsubject to x ∈ K ,y ∈ K ∗Cx = β , C T z + y = α(p)−(d) (265)2.46 A convex problem, essentially, has convex objective function optimized over a convexset. (4) In this context, (p) is convex if K is a convex cone. Slater’s condition is satisfiedwhenever any primal strictly feasible point exists. (p.235)

2.13. DUAL CONE & GENERALIZED INEQUALITY 141whose optimal objective always has the saddle value 0 (regardless of theparticular convex cone K and other problem parameters). [267,3.2] Thusdetermination of convergence for either primal or dual problem is facilitated.2.13.1.1 Key properties of dual coneFor any cone, (−K) ∗ = −K ∗For any cones K 1 and K 2 , K 1 ⊆ K 2 ⇒ K ∗ 1 ⊇ K ∗ 2 [245,2.7](Cartesian product) For closed convex cones K 1 and K 2 , theirCartesian product K = K 1 × K 2 is a closed convex cone, andK ∗ = K ∗ 1 × K ∗ 2 (266)(conjugation) [228,14] [73,4.5] When K is any convex cone, the dualof the dual cone is the closure of the original cone; K ∗∗ = K . BecauseK ∗∗∗ = K ∗ K ∗ = (K) ∗ (267)When K is closed and convex, then the dual of the dual cone is theoriginal cone; K ∗∗ = K .If any cone K has nonempty interior, then K ∗ is pointed;K nonempty interior ⇒ K ∗ pointed (268)Conversely, if the closure of any convex cone K is pointed, then K ∗ hasnonempty interior;K pointed ⇒ K ∗ nonempty interior (269)Given that a cone K ⊂ R n is closed and convex, K is pointed ifand only if K ∗ − K ∗ = R n ; id est, iff K ∗has nonempty interior.[41,3.3, exer.20](vector sum) [228, thm.3.8] For convex cones K 1 and K 2K 1 + K 2 = conv(K 1 ∪ K 2 ) (270)

2.13. DUAL CONE & GENERALIZED INEQUALITY 141whose optimal objective always has the saddle value 0 (regardless of theparticular convex cone K and other problem parameters). [267,3.2] Thusdetermination of convergence for either primal or dual problem is facilitated.2.13.1.1 Key properties of dual coneFor any cone, (−K) ∗ = −K ∗For any cones K 1 and K 2 , K 1 ⊆ K 2 ⇒ K ∗ 1 ⊇ K ∗ 2 [245,2.7](Cartesian product) For closed convex cones K 1 and K 2 , theirCartesian product K = K 1 × K 2 is a closed convex cone, andK ∗ = K ∗ 1 × K ∗ 2 (266)(conjugation) [228,14] [73,4.5] When K is any convex cone, the dualof the dual cone is the closure of the original cone; K ∗∗ = K . BecauseK ∗∗∗ = K ∗ K ∗ = (K) ∗ (267)When K is closed and convex, then the dual of the dual cone is theoriginal cone; K ∗∗ = K .If any cone K has nonempty interior, then K ∗ is pointed;K nonempty interior ⇒ K ∗ pointed (268)Conversely, if the closure of any convex cone K is pointed, then K ∗ hasnonempty interior;K pointed ⇒ K ∗ nonempty interior (269)Given that a cone K ⊂ R n is closed and convex, K is pointed ifand only if K ∗ − K ∗ = R n ; id est, iff K ∗has nonempty interior.[41,3.3, exer.20](vector sum) [228, thm.3.8] For convex cones K 1 and K 2K 1 + K 2 = conv(K 1 ∪ K 2 ) (270)

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