CW I c Gabriel Nagy Convexity “Warm-up” I: Convex Sets and Convex Hulls Notes from the Functional Analysis Course (Fall 07 - Spring 08) Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. In this section, however, when K = C, it will be only the real linear structure that will play a significant role. Definition. Suppose X is a vector space. A subset C ⊂ X is said to be convex, if: • whenever a, b ∈ C, and τ ∈ [0, 1], it follows that τa + (1 − τ)b ∈ C. Exercises 1-7. T 1. Prove that, if (Ci)i∈I is a collection of convex subsets of X , then the intersection i∈I Ci is also convex. 2. Prove that, given a collection of vector spaces (Xi)i∈I and for each i ∈ I a convex subset Q Q Ci ⊂ Xi, the product set C = i∈I Ci is convex in the product space X = i∈I Xi. 3. Suppose X and Y are vector spaces, and φ : X → Y is linear. Prove that (i) if C ⊂ X is convex, then φ(C) ⊂ Y is convex; (i) if D ⊂ Y is convex, then φ−1(D) ⊂ X is convex. 4. Prove that if A, B ⊂ X are convex, then so is A + B. 5. Prove that, if C is convex, then αC + βC = (α + β)C, ∀ α, β ≥ 0. 6. Suppose n ∈ N. Show that the set n ∆n = {(τ1, . , τn) ∈ [0, 1] : τ1 + ··· + τn = 1} is convex in Rn. This set is referred to as the standard (n − 1)-simplex, and is often used in Algebraic Topology. 7. Suppose C is convex. Prove that for any n ∈ N, for any x1, . , xn ∈ C, and for any (τ1, . , τn) ∈ ∆n, the point x = τ1x1 + ··· + τnxn also belongs to C. 1 Proposition-Definition Let X be a vector space, and let A ⊂ X be a non-empty subset. Using the notations from Exercise 5, define for each n ∈ N, the set Sn(A) = {x ∈ X : ∃(τ1, . , τn) ∈ ∆n, ∃a1, . , an ∈ A, s.t. x = τ1a1 + ··· + τnan} . S∞ The set conv(A) = n=1 Sn(A) has the following properties: (i) conv(A) ⊃ A; (ii) conv(A) is convex; (iii) whenever C is some convex set, such that C ⊃ A, it follows that C ⊃ conv(A). The set conv(A) is called the convex hull of A. Proof. Part (i) is trivial, since S1(A) ⊃ A (write any element a ∈ A as 0a + 1a). To prove assertion (ii), we simply notice that we have the inclusions τSm(A) + (1 − τ)Sn(A) ⊂ Sm+n(A), ∀ τ ∈ [0, 1], m, n ∈ N. The assertion (iii) is also clear, since by convexity we clearly have (see Exercise 6 above) the inclusions C ⊃ Sn(A), ∀ n ∈ N. Comment. Based on the preceding discussion, the convex hull conv(A) is the smallest convex set that contains A. By Exercise 1, this can also be described as \ conv(A) = C. A⊂C⊂X C convex Exercises 8-12. 8. Use the notations from Proposition-Definition 1. Prove that, if A = C1 ∪ · · · ∪ Cn, with C1,..., Cn ⊂ X convex, then conv A = {τ1x1 + ··· + τnxn : x1 ∈ C1, . , xn ∈ Cn, (τ1, . , τn) ∈ ∆n} = Sn(A). In particular, if A ⊂ X has n elements, then conv(A) = Sn(A). 9. Prove that, if A ⊂ X is absorbing, then conv(A) is also absorbing. 10. Prove that, if A ⊂ X is balanced, then conv(A) is also balanced. 11. Suppose X is equipped with a linear topology. Prove that, if A ⊂ X is open, then conv(A) is also open. 12. Suppose X is equipped with a linear topology, and C ⊂ X is convex. Prove that the interior Int(C) and the closure C are also convex. 2 N 13*. Prove that if A ⊂ R has N + 2 elements, then conv A = SN+1(A). (Hint: Use induction on N. Assuming the result is true for N − 1, start with a set A = {a1, . , aN+1, aN+2 = 0}, and consider the set A0 = {a1, . , aN , 0}, so that if we take C0 = conv A0, we have conv A = {taN+1 + (1 − t)x : x ∈ C0, t ∈ [0, 1]}. Use the inductive hypothesis in the case when dim span A0 ≤ N −1, so that only the case when a1, . , aN are linearly independent remains to be treated. So now we may very well as- N N sume that a1, . , aN are the standard basis in R , thus C0 = {(x1, . , xN ) ∈ [0, 1] : x1 + ··· + xN ≤ 1}. Fix now some vector y ∈ C, written as y = taN+1 + (1 − t)x, for some t ∈ [0, 1] and x ∈ C0, and let us show the existence of N + 1 points in A, such that y is a convex combination of them. If y ∈ C0, there is nothing to prove. Assume then y 6∈ C0, in which case one can show that y can also be written as y = raN+1 + (1 − r)z with r ∈ [0, 1], but z on the boundary of C0. The point z is con- structed as z = s0aN+1 +(1−s0)x, where s0 = max{s ∈ [0, 1] : saN+1 +(1−s)x ∈ C0}. An obvious compactness argument is needed here. The proof is then finished by show- ing that any z on the boundary of C0 lies on a face of C0, that is, z is a convex combination of N vectors in A0.) N 14*. Show that for any non-empty set A ⊂ R , one has the equality conv A = SN+1(A). (Hint: Reduce first to the case when A is finite. Use induction on card A, and reduce to the previous Exercise.) 15. Show that, if A ⊂ Kn is compact, then conv A is also compact. (Hint: Use the previous Exercise.) 16. Give an example of a closed subset A ⊂ R2, such that conv A is not closed. (Hint: Try A = {(x, x2): x ≥ 0}.) 3.