8 I. Convex Sets at Large m = 1, then u = u1 and u ∈ A since S ⊂ A. Suppose that m 1. Then αm 1 and we may write u = (1 − αm)w + αmum, where w = α1 1 − αm u1 + . . . + αm−1 1 − αm um−1. Now, w is a convex combination of u1,... , um−1 because m−1 i=1 αi 1 − αm = 1 1 − αm m−1 i=1 αi = 1 − αm 1 − αm = 1. Therefore, by the induction hypothesis, we have w ∈ A. Since A is convex, [w, um] ⊂ A, so u ∈ A. PROBLEMS. 1◦. Prove that conv ( conv(S) ) = conv(S) for any S ⊂ V . 2◦. Prove that if A ⊂ B, then conv(A) ⊂ conv(B). 3◦. Prove that conv(A) ∪ conv(B) ⊂ conv(A ∪ B). 4. Let S ⊂ V be a set and let u, v ∈ V be points such that u / conv(S) and v / conv(S). Prove that if u ∈ conv S ∪ {v} ) and v ∈ conv ( S ∪ {u} , then u = v. 5 (Gauss-Lucas Theorem). Let f(z) be a non-constant polynomial in one com- plex variable z and let z1,... , zm be the roots of f (that is, the set of all solutions to the equation f(z) = 0). Let us interpret a complex number z = x + iy as a point (x, y) ∈ R2. Prove that each root of the derivative f (z) lies in the convex hull conv(z1, . . . , zm). Hint: Without loss of generality we may suppose that f(z) = (z − z1) · · · (z − zm). If w is a root of f (z), then ∑ m i=1 j=i (w − zj) = 0, and, therefore, ∑ m i=1 j=i (w − zj) = 0, where z is the complex conjugate of z. Multiply both sides of the last identity by (w − z1) · · · (w − zn) and express w as a convex combi- nation of z1,... , zm. Next, we introduce two important classes of convex sets. (2.2) Definitions. The convex hull of a finite set of points in Rd is called a polytope. Let c1,... , cm be vectors from Rd and let β1,... , βm be numbers. The set P = x ∈ Rd : ci,x≤ βi for i = 1, . . . , m is called a polyhedron (see Problem 2 of Section 1.3).

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