12 I. Convex Sets at Large (2.4) Corollary. If S Rd is a compact set, then conv(S) is a compact set. Proof. Let Δ Rd+1 be the standard d-dimensional simplex see Problem 1 of Section 2.2: Δ = (α1, . . . , αd+1) : d+1 i=1 αi = 1 and αi 0 for i = 1, . . . , d + 1 . Then Δ is compact and so is the direct product Sd+1 ×Δ = ( u1,... , ud+1 α1,... , αd+1 ) : ui S and (α1, . . . , αd+1) Δ . Let us consider the map Φ : Sd+1 × Δ −→ Rd, Φ(u1, . . . , ud+1 α1,... , αd+1) = α1u1 + . . . + αd+1ud+1. Theorem 2.3 implies that the image of Φ is conv(S). Since Φ is continuous, the image of Φ is compact, which completes the proof. PROBLEMS. 1. Give an example of a closed set in R2 whose convex hull is not closed. 2. Prove that the convex hull of an open set in Rd is open. 3. An Application: Positive Polynomials In this section, we demonstrate a somewhat unexpected application of Cara- th´ eodory’s Theorem (Theorem 2.3). We will use Carath´ eodory’s Theorem in the space of (homogeneous) polynomials. Let us fix positive integers k and n and let H2k,n be the real vector space of all homogeneous polynomials p(x) of degree 2k in n real variables x = (ξ1, . . . , ξn). We choose a basis of H2k,n consisting of the monomials ea = ξα1 1 · · · ξn αn for a = (α1, . . . , αn) where α1 + . . . + αn = 2k. Hence dim H2k,n = ( n+2k−1 2k ) . At this point, we are not particularly concerned with choosing the “correct” scalar product in H2k,n. Instead, we declare {ea} the orthonormal basis of H2k,n, hence identifying H2k,n = Rd with d = ( n+2k−1 2k ) . We can change variables in polynomials. (3.1) Definition. Let U : Rn −→ Rn be an orthogonal transformation and let p H2k,n be a polynomial. We define q = U(p) by q(x) = p ( U −1 x ) for x = (ξ1, . . . , ξn). Clearly, q is a homogeneous polynomial of degree 2k in ξ1,... , ξn.
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