The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz during his study of the problem of moments.

Formulation

Let E be a real vector space, F\subset E be a vector subspace, and K\subset E be a convex cone. A linear functional is called K-positive, if it takes only non-negative values on the cone K: A linear functional is called a K-positive extension of \phi, if it is identical to \phi in the domain of \phi, and also returns a value of at least 0 for all points in the cone K: In general, a K-positive linear functional on F cannot be extended to a K-positive linear functional on E. Already in two dimensions one obtains a counterexample. Let and F be the x-axis. The positive functional \phi(x,0)=x can not be extended to a positive functional on E. However, the extension exists under the additional assumption that namely for every y\in E, there exists an x\in F such that y-x\in K.

Proof

The proof is similar to the proof of the Hahn–Banach theorem (see also below). By transfinite induction or Zorn's lemma it is sufficient to consider the case dim E/F = 1. Choose any. Set We will prove below that. For now, choose any c satisfying, and set \psi(y) = c, , and then extend \psi to all of E by linearity. We need to show that \psi is K-positive. Suppose z \in K. Then either z = 0, or or for some p > 0 and x \in F. If z = 0, then \psi(z) > 0. In the first remaining case, and so by definition. Thus In the second case, x - y \in K, and so similarly by definition and so In all cases, \psi(z) > 0, and so \psi is K-positive. We now prove that. Notice by assumption there exists at least one x \in F for which y - x \in K, and so -\infty < a. However, it may be the case that there are no x \in F for which x - y \in K, in which case b = \infty and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that b < \infty and there is at least one x \in F for which x - y \in K. To prove the inequality, it suffices to show that whenever x \in F and y - x \in K, and x' \in F and, then. Indeed, since K is a convex cone, and so since \phi is K-positive.

Corollary: Krein's extension theorem

Let E be a real linear space, and let K ⊂ E be a convex cone. Let x ∈ E/(−K) be such that R x + K = E. Then there exists a K-positive linear functional φ: E → R such that φ(x) > 0.

Connection to the Hahn–Banach theorem

The Hahn–Banach theorem can be deduced from the M. Riesz extension theorem. Let V be a linear space, and let N be a sublinear function on V. Let φ be a functional on a subspace U ⊂ V that is dominated by N: The Hahn–Banach theorem asserts that φ can be extended to a linear functional on V that is dominated by N. To derive this from the M. Riesz extension theorem, define a convex cone K ⊂ R×V by Define a functional φ1 on R×U by One can see that φ1 is K-positive, and that K + (R × U) = R × V. Therefore φ1 can be extended to a K-positive functional ψ1 on R×V. Then is the desired extension of φ. Indeed, if ψ(x) > N(x), we have: (N(x), x) ∈ K, whereas leading to a contradiction.

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