Rakesh V. Vohra's Advanced mathematical economics PDF

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The figure assumes that b = x ∗ . The line labeled L is perpendicular to the segment [b, x ∗ ] and is chosen to be midway between x ∗ and b. The line L is our candidate for the straight line that separates b from C. For the line L to be our separator, we need to show that no point y ∈ C lies to the left of L. 4. 4 RAKE: “chap03” — 2004/9/17 — 06:10 — page 34 — #2 Convex sets 35 C, every point on the line segment joining y to x ∗ is also in C. 4 is in C. The point z is chosen so that the line joining b to z is perpendicular to the line joining x ∗ to y.

X r are a finite collection of vectors in a convex set C, then ri=1 λi x i is also in C where ri=1 λi = 1 and λi ≥ 0 for all i. One could just as well define convexity of a set C by requiring that the weighted average of any finite subset of points in C also be in C. Verifying convexity would require checking this condition for every finite subset of points. The definition given above says that it suffices to check every pair of points, presumably a less laborious task. Convex sets have many useful properties.

Further H is closed so, H ∩ S is closed and since S is bounded so is H ∩ S. However H ∩ S exists in dimension n − 1 and the induction hypothesis applies. So, H ∩ S has extreme points and every point in H ∩ S is in the convex hull of these extreme points. It remains to prove that every extreme point of H ∩ S is an extreme point of S, suppose not. Let x be an extreme point of H ∩ S. Since x is not an extreme point of S exists y, z ∈ S such that x = λy + (1 − λ)z for λ ∈ (0, 1). Then c = h · x = λh · y + (1 − λ)h · z ≥ c.

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