The main part of Hilbert's Theorem 6.1 is the following

Theorem 7.1. (Hilbert, 1888)Every PSD ternary quartic is a sum of squares of ternary quadratics and indeed three squares always suffice.

We plan to give two proofs in this chapter. The first proof, due to Choi and Lam [C9], uses arguments from elementary analysis – and also makes use of the Krein-Milman theorem, which is a popular tool of functional analysts. This proof only shows the first part of the theorem, i.e. that P3, 4 = Σ3, 4.

The second proof is Hilbert's original. To give the Choi-Lam proof we need the following.

Lemma 1. Let T(X, Y, Z) ∈ R3, 4. Then there exists a quadratic form q(X, Y, Z) (≠ 0) such that T ≥ q2, where by T ≥ q2 we mean of course that the form T – q2 ≥ 0 i.e. is PSD.

Proof. Let S(T) denote the set of zeros of T.

Case 1: S(T) = φ. Consider the positive continuous function

defined for all (X, Y, Z) ≠ (0, 0, 0). On the unit sphere S2 (a compact set) let μ = inf j ≥ 0. By compactness of S, μ is attained, i.e. μ = j(α, β, γ) for some (α, β, γ) ∈ S2; so μ ≠ 0 since S(T) = φ. Thus We claim that (7.1) holds for all the points of R3; for let (α, β, γ) ≠ (0, 0, 0) be any point of R3 and let.