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Convex Sets and Convex Functions (Part I)

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Convex Sets and Convex Functions (Part I) Convex Sets and Convex Functions (part I) Prof. Dan A. Simovici UMB 1 / 79 Outline 1 Convex and Affine Sets 2 The Convex and Affine Closures 3 Operations on Convex Sets 4 Cones 5 Extreme Points 2 / 79 Convex and Affine Sets n Special Subsets in R Let L be a real linear space and let x; y 2 L. The closed segment determined by x and y is the set [x; y] = f(1 − a)x + ay j 0 6 a 6 1g: The half-closed segments determined by x and y are the sets [x; y) = f(1 − a)x + ay j 0 6 a < 1g; and (x; y] = f(1 − a)x + ay j 0 < a 6 1g: The open segment determined by x and y is (x; y) = f(1 − a)x + ay j 0 < a < 1g: The line determined by x and y is the set `x;y = f(1 − a)x + ay j a 2 Rg: 3 / 79 Convex and Affine Sets Definition A subset C of L is convex if we have [x; y] ⊆ C for all x; y 2 C. Note that the empty subset and every singleton fxg of L are convex. 4 / 79 Convex and Affine Sets Convex vs. Non-convex x3 x4 x2 x4 x x y x2 x3 y x1 x1 (a) (b) 5 / 79 Convex and Affine Sets Example The set n of all vectors of n having non-negative components is a R>0 R n convex set called the non-negative orthant of R . 6 / 79 Convex and Affine Sets Example The convex subsets of (R; +; ·) are the intervals of R. Regular polygons 2 are convex subsets of R . Example Every linear subspace T of a real linear space L is convex. 7 / 79 Convex and Affine Sets Example Let (L; k · k) be a normed linear space. An open sphere B(x0; r) ⊆ L is convex. Indeed, suppose that x; y 2 B(x0; r), that is, k x − x0 k< r and k x0 − y k< r. Let a 2 [0; 1] and let z = (1 − a)x + ay. We have k x0 − z k = k x0 − (1 − a)x − ay k = k a(x0 − y) + (1 − a)(x0 − x) k 6 a k x0 − y k +(1 − a) k x0 − x k< r: so z 2 B(x0; r). Similarly, a closed sphere B[x0; r] is a convex set. 8 / 79 Convex and Affine Sets Theorem Let x; y; z be three distinct points in the real linear space L such that z 2 `x;y . Then, one of these points belongs to the open segment determined by the remaining two points. 9 / 79 Convex and Affine Sets Proof Since z 2 `x;y , we have z = (1 − a)x + ay for some a 2 R. ¨ y ¨¨ ¨ a > 1 ¨¨ x ¨¨ ¨ u ¨¨ 0 < a < 1 ¨ ¨ a <u0 We have a 62 f0; 1g because the points x; y; z are distinct. If a > 1 we a−1 1 a−1 1 have y = a x + a z, so y 2 (x; z) because a ; a 2 (0; 1). If 0 < a < 1 a −a we have z 2 (x; y). Finally, if a < 0, since x = 1 + 1−a z + 1−a y, we have x 2 (z; y). 10 / 79 Convex and Affine Sets Definition Let U be a subset of a real linear space L and let x1;:::; xk 2 U. A linear combination of U, a1x1 + ··· + ak xk , where a1;:::; ak 2 R and k > 1 is: Pk an affine combination of U if i=1 ai = 1; a non-negative combination of U if ai > 0 for 1 6 i 6 k; a positive combination of U if ai > 0 for 1 6 i 6 k; a convex combination of U if it is both a non-negative and an affine combination of U. 11 / 79 Convex and Affine Sets Theorem Let L be a real linear space. A subset C of L is convex if and only if any convex combination of elements of C belongs to C. 12 / 79 Convex and Affine Sets Proof The sufficiency of this condition is immediate. To prove its necessity consider x1;:::; xk 2 C and the convex combination y = a1x1 + ··· + ak xk : We prove by induction on k > 1 that y 2 C. The base case, m = 1 is immediate since in this case y = a1x1 and a1 = 1. For the inductive step, suppose that the statement holds for k and let y be given by y = a1x1 + ··· + ak xk + ak+1xk+1, where a1 + ··· + ak + ak+1 = 1, ai > 0 and xi 2 C for 1 6 i 6 k + 1. We have k X ai y = (1 − ak+1) xi + ak+1xk+1: 1 − ak+1 i=1 Pk ai Since z = xi is a convex combination of k vectors, we have i=1 1−ak+1 z 2 C by the inductive hypothesis, and the equality y = (1 − ak+1)z + ak+1xk+1 implies y 2 C. 13 / 79 Convex and Affine Sets Definition Let L; K be two linear spaces. A mapping f : L −! K is affine when there exists a linear mapping h : L −! K and some b 2 K such that f (x) = h(x) + b for every x 2 L 14 / 79 Convex and Affine Sets Theorem Let L; K be two linear spaces and let f : L −! K be an affine mapping. If C be a convex subset of L, then f (C) is a convex subset of K. If D is a convex subset of K, then f −1(D) = fx 2 L j f (x) 2 Dg is a convex subset of L. 15 / 79 Convex and Affine Sets Proof Since f is an affine mapping, we have f (x) = h(x) + b, where h : L −! K is a linear mapping and b 2 K for x 2 L. Therefore, if y1; y2 2 f (C) we can write y1 = h(x1) + b and y2 = h(x2) + b. This, in turn, allows us to write for a 2 [0; 1]: (1 − a)y1 + ay2 = (1 − a)h(x1) + (1 − a)b + ah(x2) + ab = h((1 − a)x1 + ax2) + b = f ((1 − a)x1 + ax2): The convexity of C implies (1 − a)x1 + ax2 2 C, so (1 − a)y1 + ay2 2 f (C), which shows that f (C) is convex. 16 / 79 Convex and Affine Sets Definition A subset C of a linear space L is affine subspace if `x;y ⊆ C for all x; y 2 C. In other words, C is a non-empty affine subspace if every point on the line determined by two members of C, x and y belongs to C. Note that C is a subspace of L if and only if 0L 2 C and C is an affine subspace. 17 / 79 Convex and Affine Sets Example The empty set ;, every singleton fxg, and the entire space L are affine subspaces of L. Also, every hyperplane H is an affine subspace of L. Theorem A non-empty subset C of a linear space L is an affine subspace if and only if any affine combination of elements of C belongs to C. 18 / 79 Convex and Affine Sets It is immediate to verify that any translation of a linear space K is an affine subspace. The converse is also true as we show next. Theorem Let D be a non-empty affine subspace in a linear space L. There exists a translation tu and a unique subspace K of L such that D = tu(K). 19 / 79 Convex and Affine Sets Proof Let K = fx − y j x; y 2 Dg and let x0 2 D. We have 0L = x0 − x0 2 K and it is immediate that K is subspace of L. Let u be an element of D. We claim that D = tu(K). Indeed, if z 2 D, z − u 2 K, so z 2 tu(K), which implies D ⊆ tu(K). Conversely, if x 2 tu(K) we have x = u + v for some v 2 K and, therefore, x = u + s − t for some s; t 2 D, where v = s − t. This implies x 2 D because u + s − t is an affine combination of D. 20 / 79 Convex and Affine Sets Proof (cont'd) To prove the uniqueness of the subspace K suppose that D = tu(K1) = tv(K2), where both K1 and K2 are subspaces of L. Since 0L 2 K2, it follows that there exists w 2 K1 such that u + w = v. Similarly, since 0L 2 K1, it follows that there exists t 2 K1 such that u = v + t, which implies w + t = 0L. Thus, both w and t belong to both subspaces K1 and K2. If s 2 K1, it follows that u + s = v + z for some z 2 K2. Therefore, s = (v − u) + z 2 K2 because w = v − u 2 K2. This implies K1 ⊆ K2. The reverse inclusion can be shown similarly. 21 / 79 Convex and Affine Sets Definition Let D be a non-empty affine subspace in a linear space L. The dimension of D (denoted by dim(D)) is the dimension of the unique subspace K of L such that D = tu(K) for some translation tu of L. The dimension of a convex set C is the dimension of the affine space Kaff(C). 22 / 79 Convex and Affine Sets Since ; is an affine subspace of L and there is no subspace of L that can be translated into ;, the dimension of ; is set through the special definition dim(;) = −1.
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