3. An Application: Positive Polynomials 13 PROBLEMS. 1◦. Check that (U1U2)(p) = U1(U2(p)). 2◦. Let p(x) = x 2k = ( ξ2 1 + . . . + ξ2 n )k . Prove that U(p) = p for any orthogonal transformation U. It turns out that the polynomial of Problem 2, Section 3.1, up to a scalar multiple, is the only polynomial that stays invariant under any orthogonal trans- formation. (3.2) Lemma. Let p ∈ H2k,n be a polynomial such that U(p) = p for every or- thogonal transformation U. Then p(x) = γ x 2k = γ ( ξ1 2 + . . . + ξn 2 )k for some γ ∈ R. Proof. Let us choose a point y ∈ Rd such that y = 1 and let γ = p(y). Let us consider q(x) = p(x) − γ x 2k . Thus q is a homogeneous polynomial of degree 2k and q(Ux) = q(x) for any or- thogonal transformation U and any vector x. Moreover, q(y) = 0. Since for every vector x such that x = 1 there is an orthogonal transformation Ux such that Uxy = x, we have q(x) = q ( Uxy ) = q(y) = 0 and hence q(x) = 0 for all x such that x = 1. Since q is a homogeneous polynomial, we have q(x) = 0 for all x ∈ Rn. Therefore, p(x) = γ x 2k as claimed. We are going to use Theorem 2.3 to deduce the existence of an interesting identity. (3.3) Proposition. Let k and n be positive integers. Then there exist vectors c1,... , cm ∈ Rn such that x 2k = m i=1 ci,x 2k for all x ∈ Rn. In words: the k-th power of the sum of squares of n real variables is a sum of 2k-th powers of linear forms in the variables. Proof. We are going to apply Carath´ eodory’s Theorem in the space H2k,n. Let Sn−1 = c ∈ Rn : c = 1 be the unit sphere in Rn. For a c ∈ Sn−1, let pc(x) = c, x 2k where x = (ξ1, . . . , ξn).

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2002 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.