3. An Application: Positive Polynomials 15 3. Prove that one can choose m ( n+2k−1 2k ) in Proposition 3.3. Remark: In his solution of Waring’s problem, for all positive integers k and n, D. Hilbert constructed integer vectors ci and rational numbers γi such that x 2k = m i=1 γi ci,x 2k for all x Rn see, for example, Chapter 3 of [N96]. We apply Proposition 3.3 to study positive polynomials. (3.4) Definition. Let p H2k,n be a polynomial. We say that p is positive provided p(x) 0 for all x = 0. Equivalently, p H2k,n is positive provided p(x) 0 for all x Sn−1. Similarly, a polynomial p H2k,n is non-negative if p(x) 0 for all x. PROBLEM. 1◦. Prove that the set of all positive polynomials is a non-empty open convex set in H2k,n and that the set of all non-negative polynomials is a non-empty closed convex set in H2k,n. We apply Proposition 3.3 to prove that a homogeneous polynomial is positive if and only if it can be multiplied by a sufficiently high power of x 2 to produce a sum of even powers of linear functions. The proof below is due to B. Reznick [R95] and [R00]. (3.5) Proposition. Let p H2k,n be a positive polynomial. Then there exist a positive integer s and vectors c1,... , cm Rn such that x 2s−2k p(x) = m i=1 ci,x 2s for all x Rn. Sketch of Proof. For a polynomial f H2k,n, f(x) = a=(α1,... ,αn) λaξ1 α1 . . . ξαn, n let us formally define the differential operator f(∂) = a=(α1,... ,αn) λa ∂α1 ∂ξα1 1 · · · ∂αn ∂ξn αn . Let us choose a positive integer s 2k and the corresponding identity of Proposition 3.3: (3.5.1) x 2s = m i=1 ci,x 2s .
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