14 I. Convex Sets at Large Hence we have pc H2k,n. Let K = conv pc : c Sn−1 be the convex hull of all polynomials pc. Since the sphere Sn−1 is compact and the map c −→ pc is continuous, the set pc : c Sn−1 is a compact subset of H2k,n. Therefore, by Corollary 2.4, we conclude that K is compact. Let us prove that γ x 2k K for some γ 0. The idea is to average the polynomials pc over all possible vectors c Sn−1. To this end, let dc be the rotation invariant probability measure on Sn−1 and let (3.3.1) p(x) = Sn−1 pc(x) dc = Sn−1 c, x 2k dc be the average of all polynomials pc. We observe that p H2k,n. Moreover, since dc is a rotation invariant measure, we have U(p) = p for any orthogonal transformation U of Rn and hence by Lemma 3.2, we must have p(x) = γ x 2k for some γ R. We observe that γ 0. Indeed, for any x = 0, we have pc(x) 0 for all c Sn−1 except from a set of measure 0 and hence p(x) 0. The integral (3.3.1) can be approximated with arbitrary precision by a finite Riemann sum: p(x) 1 N N i=1 pc i (x) for some ci Sn−1. Therefore, p lies in the closure of K. Since K is closed, p K. By Theorem 2.3, we can write p(x) = γ x 2k as a convex combination of some ( n+2k−1 2k ) +1 polynomials pc i (x) = ci,x 2k . Dividing by γ, we complete the proof. It is not always easy to come up with a particular choice of ci in the identity of Proposition 3.3. PROBLEMS. 1. Prove Liouville’s identity: (ξ2 1 + ξ2 2 + ξ2 3 + ξ2)2 4 = 1 6 1≤ij≤4 (ξi + ξj)4 + 1 6 1≤ij≤4 (ξi ξj)4. 2. Prove Fleck’s identity: (ξ1 2 +ξ2 2 +ξ3 2 +ξ4)3 2 = 1 60 1≤ijk≤4 ( ξi ±ξj ±ξk )6 + 1 30 1≤ij≤4 ( ξi ±ξj )6 + 3 5 1≤i≤4 ξi 6 , where the sums containing ± signs are taken over all possible independent choices of pluses and minuses.
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