2006 Recent Developments in Nonlinear Optimization Theory (slide).pdf

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2006 Recent Developments in Nonlinear Optimization Theory (slide).pdf

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2006 Recent Developments in Nonlinear Optimization Theory (slide)

NUS Graduate University of Chinese Academy of Sciences 1 $ % Recent Developments in Nonlinear Optimization Theory Defeng Sun Department of Mathematics National University of Singapore Republic of Singapore July 9-11, 2006 NUS Graduate University of Chinese Academy of Sciences 2 $ % 2 Variational Analysis on Metric Projectors Over Closed Convex Sets Let Z be a finite-dimensional Hilbert vector space equipped with a scalar product 〈·, ·〉 and its induced norm ‖ · ‖ and D be a nonempty closed convex set in Z. For any z ∈ Z, let ΠD(z) denote the metric projection of z onto D: min 1 2 〈y ? z, y ? z〉 s.t. y ∈ D. (1) The operator ΠD : Z → Z is called the metric projection operator or metric projector over D. NUS Graduate University of Chinese Academy of Sciences 3 $ % Proposition 2.1 Let D be a nonempty closed convex set in Z. Then the point y ∈ D is an optimal solution to (1) if and only if it satisfies 〈z ? y, d? y〉 ≤ 0 ? d ∈ D . (2) Proof. “?” Suppose that y ∈ D is an optimal solution to (1). Let d be an arbitrary point in D. Then yt := (1? t)y + td ∈ D for any t ∈ [0, 1]. This, together with the fact that y is an optimal solution, implies that ‖z ? yt‖2 ≥ ‖z ? y‖2 ? t ∈ [0, 1], which further implies ‖(1? t)(z ? y) + t(z ? d)‖2 ≥ ‖z ? y‖2 ? t ∈ [0, 1]. NUS Graduate University of Chinese Academy of Sciences 4 $ % Thus, (t2? 2t)‖z? y‖2 + 2t(1? t)〈z? y, z? d〉+ t2‖z? d‖2 ≥ 0 ? t ∈ [0, 1]. By taking t ↓ 0 and dividing t on both sides of the above equation, we obtain ?2‖z ? y‖2 + 2〈z ? y, z ? d〉 ≥ 0 , which turns into (2). “?” Suppose that y ∈ D satisfies (2). Assume on the contrary that y does not solve (1). Then we have by the assumption, 〈z ? y, ΠD(z)? y〉 ≤ 0 and by the sufficiency part, 〈z ?ΠD(z), y ?ΠD(z)〉 ≤ 0 . NUS Graduate University of Chinese Academy of Sciences 5 $ % Summing up the above two inequalities leads to 〈ΠD(z)? y, ΠD(z)? y〉 ≤ 0 . This implies that y = ΠD(z). The contradiction shows that y solves (1). ¤ Note that Proposition 2.1 holds even if Z is infin