4. Convex Sets and (Quasi-)Concave Functions Daisuke Oyama - PowerPoint PPT Presentation

4. Convex Sets and (Quasi-)Concave Functions Daisuke Oyama Mathematics II April 17, 2020

Convex Sets Definition 4.1 ▶ A ⊂ R N is convex if (1 − α ) x + αx ′ ∈ A whenever x, x ′ ∈ A and α ∈ [0 , 1] . ▶ A ⊂ R N is strictly convex if (1 − α ) x + αx ′ ∈ Int A whenever x, x ′ ∈ A , x ̸ = x ′ , and α ∈ (0 , 1) . 1 / 39

Convex Combinations ▶ For α 1 , . . . , α M ≥ 0 , ∑ M m =1 α m = 1 , α 1 x 1 + · · · + α M x M is called a convex combination of x 1 , . . . , x M . Proposition 4.1 If A ⊂ R N is convex, then any convex combination of elements in A is contained in A . Proof By induction. 2 / 39

Convex Hull Proposition 4.2 For any index set Λ , if C λ ⊂ R N is convex for all λ ∈ Λ , then ∩ λ ∈ Λ C λ is convex. (The intersection of any family of convex sets is convex.) Definition 4.2 For A ⊂ R N , the convex hull of A , denoted Co A , is the intersection of all convex sets that contain A (or, the smallest convex set that contains A ). 3 / 39

Proposition 4.3 For A ⊂ R N , Co A equals the set of all convex combinations of elements in A , i.e., M { � ∑ x ∈ R N � α m x m Co A = � x = � m =1 for some M ∈ N , x 1 , . . . , x M ∈ A , and M } ∑ α 1 , . . . , α M ≥ 0 with α m = 1 . m =1 4 / 39

Proposition 4.4 Let A, B ⊂ R N . 1. A ⊂ Co A . 2. If A ⊂ B , then Co A ⊂ Co B . 3. Co Co A = Co A . 5 / 39

Algebra of Convex Sets Proposition 4.5 1. If A, B ⊂ R N are convex, then A + B = { x ∈ R N | x = a + b for some a ∈ A and b ∈ B } is convex. 2. If A ⊂ R N is convex, then for t ∈ R , tA = { x ∈ R N | x = ta for some a ∈ A } is convex. 6 / 39

Proposition 4.6 M M ∑ ∑ For A 1 , . . . , A M ⊂ R N , Co A m = Co A m . m =1 m =1 Proof ▶ ( LHS ) ⊂ ( RHS ) : Exercise. ▶ ( LHS ) ⊃ ( RHS ) : Sufficient to show for M = 2 : If x ∈ Co A 1 + Co A 2 , then for some y 1 , . . . , y I ∈ A 1 and z 1 , . . . , z J ∈ A 2 , we have i α i y i + ∑ j β j z j = ∑ j β j ( y i + z j ) , x = ∑ ∑ i α i where α i ≥ 0 , β j ≥ 0 , and ∑ i α i = ∑ j β j = 1 . This implies that x ∈ Co Co( A 1 + A 2 ) = Co( A 1 + A 2 ) . 7 / 39

Convex Cones Definition 4.3 ▶ A ⊂ R N is a cone if x ∈ A ⇒ αx ∈ A for any α ≥ 0 . ▶ A ⊂ R N is a convex cone if x, y ∈ A ⇒ αx + βy ∈ A for any α, β ≥ 0 . (Some textbooks define with “for any α > 0 ” and “for any α, β > 0 ”.) 8 / 39

Carath´ eodory’s Theorem For A = { x 1 , . . . , x M } ⊂ R N , m =1 α m x m ∈ R N | α 1 , . . . , α M ≥ 0 } , ▶ write Cone A = { ∑ M ▶ where ∑ M m =1 α m x m , α m ≥ 0 , is called a conic combination of x 1 , . . . , x M ∈ R N . Proposition 4.7 (Carath´ eodory’s Theorem) 1. For A = { x 1 , . . . , x M } ⊂ R N , A ̸ = { 0 } , each x ∈ Cone A is written as a conic combination of linearly independent elements of A . 2. For A ⊂ R N , each x ∈ Co A is written as a convex combination of at most N + 1 elements in A . 9 / 39

Proof 1. ▶ Let x = α 1 x 1 + · · · + α M x M , (1) where we assume without loss of generality that α 1 , . . . , α M > 0 . If x 1 , . . . , x M are linearly independent, we are done. ▶ Suppose that x 1 , . . . , x M are linearly dependent, so that c 1 x 1 + · · · + c M x M = 0 (2) for some ( c 1 , . . . , c M ) ̸ = (0 , . . . , 0) . Assume that c m > 0 for some m (if c m ≤ 0 for all m , then multiply both sides by − 1 ) . { } � c m > 0 ▶ Let µ = min α m � > 0 . c m 10 / 39

Proof ▶ By (1) and (2) we have x = ( α 1 − µc 1 ) x 1 + · · · + ( α M − µc M ) x M , where ▶ α m − µc m ≥ 0 for all m , and ▶ α m − µc m = 0 for some m . Thus x has been written as a conic combination of M − 1 (or fewer) elements of { x 1 , . . . , x M } . ▶ If these M − 1 (or fewer) vectors are linearly independent, we are done. If not, repeat the same procedure. ▶ With finitely many steps, we have a conic representation with linearly independent vectors. 11 / 39

Proof 2. Follows from Part 1 and the following lemma with I = 1 . 12 / 39

Lemma 4.8 Let A 1 , . . . , A I ⊂ R N . i =1 A i , then there exist x ij ∈ A i , i = 1 , . . . , I , If x ∈ Co ∑ I j = 1 , . . . , K i , where K i ≥ 1 , such that I ∑ Co { x i 1 , . . . , x iK i } x = i =1 and I ∑ K i ≤ N + I. i =1 13 / 39

Proof ▶ Let A 1 , . . . , A I ⊂ R N , and let x ∈ Co ∑ I i =1 A i . Since Co ∑ I i =1 A i = ∑ I i =1 Co A i by Proposition 4.6, i =1 y i for some y i ∈ A i , i = 1 , . . . , I , x is written as x = ∑ I where each y i is written as y i = ∑ J i j =1 α ij y ij for some y ij ∈ A i and α ij ≥ 0 , j = 1 , . . . , J i , with ∑ J i j =1 α ij = 1 . ▶ Consider the following vectors in R N + I : z = ( x, 1 , 1 , . . . , 1 , 1) , z 1 j = ( y 1 j , 1 , 0 , . . . , 0 , 0) , j = 1 , . . . , J 1 , z 2 j = ( y 2 j , 0 , 1 , . . . , 0 , 0) , j = 1 , . . . , J 2 , . . . z Ij = ( y Ij , 0 , 0 , . . . , 0 , 1) , j = 1 , . . . , J I . ▶ By construction, z is written as a conic combination of z ij ’s: z = ∑ I ∑ J i j =1 α ij z ij . i =1 14 / 39

Proof ▶ By the cone version of Carath´ eodory’s Theorem (Proposition 4.7(1)), there are at most N + I linearly independent elements of { z ij , i = 1 , . . . , I, j = 1 , . . . , J i } such that z is written as a conic combination of them: i.e., there exist β ij ≥ 0 , i = 1 , . . . , I , j = 1 , . . . , J i , such that j =1 β ij z ij and z = ∑ I ∑ J i i =1 ∑ I i =1 |{ j = 1 , . . . , J i | β ij > 0 }| ≤ N + I . ▶ From the 1 st through N th coordinates we have x = ∑ I ∑ J i j =1 β ij y ij . i =1 ▶ From the ( N + 1) st through ( N + I ) th coordinates we have ∑ J i j =1 β ij = 1 , i = 1 , . . . , I , where β ij > 0 for at least one j . 15 / 39

Shapley-Folkman Theorem Proposition 4.9 Let A 1 , . . . , A I ⊂ R N . If x ∈ Co ∑ I i =1 A i , then ∑ ∑ x ∈ A i + Co A i i ∈I ′ i ∈{ 1 ,...,I }\I ′ for some I ′ ⊂ { 1 , . . . , I } with |I ′ | ≥ I − N . (See Kreps, Chapter 13 for an application of this theorem.) 16 / 39

Proof ▶ Let x ∈ Co ∑ I i =1 A i . ▶ Then by Lemma 4.8, there exist x ij ∈ A i , i = 1 , . . . , I , j = 1 , . . . , K i , where K i ≥ 1 , such that ▶ x = ∑ I i =1 Co { x i 1 , . . . , x iK i } , and ▶ ∑ I i =1 K i ≤ N + I . ▶ Let I ′ = { i = 1 , . . . , I | K i = 1 } , and let |I ′ | = n . ▶ Then ∑ I i =1 K i ≥ n + 2( I − n ) = 2 I − n . ▶ With ∑ I i =1 K i ≤ N + I , this implies that n ≥ I − N . 17 / 39

Topological Properties of Convex Sets Proposition 4.10 If A ⊂ R N is open, then Co A is open. Proposition 4.11 If A ⊂ R N is convex, then Int A is convex. Proposition 4.12 If A ⊂ R N is convex, then Cl A is convex. Proof ▶ Cl A = ∩ ε> 0 ( A + B ε (0)) . 18 / 39

Topological Properties of Convex Sets Fact 1 Let A ⊂ R N be a convex set. If Int(Cl A ) ̸ = ∅ , then Int A ̸ = ∅ . Proposition 4.13 Let A ⊂ R N be a convex set. Then Int(Cl A ) = Int A . 19 / 39

Topological Properties of Convex Sets Proposition 4.14 If A ⊂ R N is bounded, then Cl(Co A ) = Co(Cl A ) . In particular, if A is compact, then Co A is compact. 20 / 39

Proof ▶ Since Co A ⊃ A , we have Cl(Co A ) ⊃ Cl A . Since Cl(Co A ) is convex (Proposition 4.12), we have Cl(Co A ) ⊃ Co(Cl A ) . ▶ Since A ⊂ Cl A , we have Co A ⊂ Co(Cl A ) . We want to show that Co(Cl A ) is closed if A is bounded. ▶ Let { x m } ⊂ Co(Cl A ) , and assume x m → ¯ x . eodory’s Theorem (Proposition 4.7(2)), each x m is ▶ By Carath´ written as x m = α m 1 x m, 1 + · · · + α m N +1 x m,N +1 , where N +1 ) ∈ ∆ = { α ∈ R N +1 | α n ≥ 0 , ∑ ▶ ( α m 1 , . . . , α m n α n = 1 } , ▶ x m, 1 , . . . , x m,N +1 ∈ Cl A . 21 / 39

Proof ▶ Since ∆ and Cl A are compact, there exists a sequence α n = lim k →∞ α m ( k ) { m ( k ) } such that the limits ¯ and n x n = lim k →∞ x m ( k ) ,n exist where (¯ α N +1 ) ∈ ∆ and ¯ α 1 , . . . , ¯ x N +1 ∈ Cl A . x 1 , . . . , ¯ ¯ ▶ Hence, x 1 + · · · + ¯ x N +1 , x = ¯ ¯ α 1 ¯ α N +1 ¯ so that ¯ x ∈ Co(Cl A ) . 22 / 39

Concave Functions Definition 4.4 Let X ⊂ R N be a non-empty convex set. ▶ A function f : X → R is concave if f ((1 − α ) x + αx ′ ) ≥ (1 − α ) f ( x ) + αf ( x ′ ) for all x, x ′ ∈ X and all α ∈ [0 , 1] . ▶ f : X → R is strictly concave if f ((1 − α ) x + αx ′ ) > (1 − α ) f ( x ) + αf ( x ′ ) for all x, x ′ ∈ X with x ̸ = x ′ and all α ∈ (0 , 1) . ▶ f : X → R is convex ( strictly convex , resp.) if − f is concave (strictly concave, resp.). 23 / 39

Hypograph and Epigraph Let X ⊂ R N be a non-empty set. ▶ The hypograph of a function f : X → R is the set hyp f = { ( x, y ) ∈ R N × R | x ∈ X, y ≤ f ( x ) } . ▶ The epigraph of a function f : X → R is the set epi f = { ( x, y ) ∈ R N × R | x ∈ X, y ≥ f ( x ) } . Proposition 4.15 Let X ⊂ R N be a nonempty convex set. f : X → R is a concave (convex, resp.) function if and only if hyp f ( epi f , resp.) is a convex set. 24 / 39