16 I. Convex Sets at Large Let us see what happens if we apply f(∂) to both sides of the identity. It is not very hard to see that (3.5.2) f(∂) c, x 2s = (2s)! (2s 2k)! f(c) · c, x 2s−2k . It suffices to check the identity when f is a monomial and then it is straightforward. One can also see that (3.5.3) f(∂) x 2s = 22ks! (s 2k)! g(x) · x 2s−2k for some g H2k,n. The correspondence f −→ g defines a linear transformation Φs : H2k,n −→ H2k,n and the crucial observation is that Φs converges to the identity operator I as s grows. Again, it suffices to check this when f is a monomial, in which case Φs(f) = f + O(1/s) by the repeated application of the chain rule. Since I−1 = I, for all sufficiently large s the operator Φs is invertible and Φ−1 s converges to the identity operator I as s grows. Now we note that the set of positive polynomials is open see Problem 1 of Section 3.4. Therefore, for a sufficiently large s the polynomial q = Φ−1(p) s lies in a sufficiently small neighborhood of p = I(p) and hence is positive. Applying q(∂) to both sides of (3.5.1), by (3.5.2) and (3.5.3) we get 22ks! (s 2k)! Φs(q) · x 2s−2k = (2s)! (2s 2k)! m i=1 q(ci) ci,x 2s−2k . Now Φs(q) = p and q(ci) 0 for i = 1, . . . , m. Rescaling, we obtain a representa- tion of p · x 2s−2k as a sum of powers of linear forms. PROBLEMS. 1◦. Check formulas (3.5.2) and (3.5.3). 2◦. Check that Φs indeed converges to the identity operator on H2k,n as s grows. 3. For polynomials f, g H2k,n, let us define f, g = f(∂)g. Note that since deg f = deg g, we get a number. Prove that f, g is a scalar product in H2k,n and that U(f),U(g) = f, g for every orthogonal transformation of Rn. 4. Construct an example of a non-negative polynomial p H2k,n for which the conclusion of Proposition 3.5 does not hold true. 5. Using Proposition 3.5, deduce Polya’s Theorem:
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