Lecture 3

# Take a sequence yk c such that x yk then by simple

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Unformatted text preview: ∈ RN has a unique closest point PC (x) ∈ cl C , called the projection of x on cl C . Furthermore, for any z ∈ C we have ⟨x − PC (x), z − PC (x)⟩ ≤ 0. Proof. Let δ = inf {∥x − y ∥ | y ∈ C } ≥ 0 be the distance from x to C . Take a sequence {yk } ⊂ C such that ∥x − yk ∥ → δ . Then by simple algebra we get 1 ∥yk − yl ∥2 = 2 ∥x − yk ∥2 + 2 ∥x − yl ∥2 − 4 x − (yk + yl ) 2 2 . Since C is convex, we have 1 (yk + yl ) ∈ C , so by the deﬁnition of δ we get 2 ∥yk − yl ∥2 ≤ 2 ∥x − yk ∥2 + 2 ∥x − yl ∥2 − 4δ 2 → 2δ 2 + 2δ 2 − 4δ 2 = 0 as k, l → ∞. Since {yk } ⊂ C is Cauchy, it converges to some point y ∈ cl C . Then ∥x − y ∥ ≤ ∥x − yk ∥ + ∥yk − y ∥ → δ + 0 = δ, so y is the closest point to x in cl C . If y1 , y2 are two closest points, then by the same argument we get ∥y1 − y2 ∥2 ≤ 2 ∥x − y1 ∥2 + 2 ∥x − y2 ∥2 − 4δ 2 ≤ 0, so y1 = y2 . Thus y = PC (x) is unique. Finally, let z ∈ C be any point. Take {yk } ⊂ C such that yk → y = PC (x). Since C is convex, for any 0 < α ≤ 1 we have (1 − α)yk + αz ∈ C . Therefore 2 2 δ 2 = ∥x − y ∥ ≤ ∥x − (1 − α)yk − αz ∥ . 2 2 Letting k → ∞ we get ∥x − y ∥ ≤ ∥x − y − α(z − y )∥ . Expanding both sides, dividing by α > 0, and letting α → 0, we get ⟨x − y, z − y ⟩ ≤ 0, which is the desired inequality. If C and D don’t overlap ( no intersect) then there exists a hyperplane that separates them. IF they are BOTH closed and ONE is compact, they are STRICTLY separated 2014W Econ 172B Operations Research (B) Alexis Akira Toda The following proposition shows that a point that is not an interior point of a convex C can be separated from C . Proposition 4. Let C ⊂ RN be nonempty and convex and x ∈ int C . Then ¯/ there exists a hyperplane ⟨a, x⟩ = c that separates x and C , i.e., ¯ ⟨a, x⟩ ≥ c ≥ ⟨a, z ⟩ ¯ for any z ∈ C . If x ∈ cl C , then the above inequalities can be made strict. ¯/ Proof. Suppose that x ∈ cl C . Let y = PC (¯) be the projection of x on cl C . ¯/ x ¯ 2 1 Then x ̸= y . Let a = x − y ̸= 0 and c = ⟨a, y ⟩ + 2 ∥a∥ . Then for any z ∈ C we ¯ ¯ have ⟨x − y, z − y ⟩ ≤ 0 =⇒ ⟨a, z ⟩ ≤ ⟨a, y ⟩ < ⟨a, y ⟩ + ¯ ⟨a, x⟩ − c = ⟨x − y, x − y ⟩ − ¯ ¯ ¯ 1 2 ∥a∥ = c, 2 1 1 2 2 ∥a∥ = ∥a∥ > 0 ⇐⇒ ⟨a, x⟩ > c. ¯ 2 2 Therefore the hyperplane ⟨a, x⟩ = c strictly separates x and C . ¯ If x ∈ cl C , since x ∈ int C we can take a sequence {xk } such that xk ∈ cl C ¯ ¯/ / and xk →...
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## This document was uploaded on 02/18/2014 for the course ECON 172b at UCSD.

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