Let E be a real vector space, F ⊂ E a vector subspace, and letK ⊂ E be a convex cone. A linear functionalφ: F → R is called K-positive, if it takes only non-negative values on the coneK: A linear functional ψ: E → R is called a K-positive extension of φ, if it is identical toφ in the domain of φ, 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 -positive linear functional on E. Already in two dimensions one obtains a counterexample taking K to be the upper half plane with the open negative x-axis removed. If F is the x-axis, then the positive functionalφ = x can not be extended to a positive functional on the plane. However, the extension exists under the additional assumption that for every y ∈ Ethere existsx∈F such that y − x ∈K; in other words, if E = K + F.
Proof
The proof is similar to the proof of the Hanh-Banach theorem. By transfinite induction or Zorn's lemma it is sufficient to consider the case dim E/F = 1. Choose any y ∈ E\F. Set We will prove below that -∞ < a ≤ b. For now, choose any c satisfying a ≤ c ≤ b, and set ψ = c, ψ|F = φ, and then extend ψ to all of E by linearity. We need toshow that ψ is K-positive. Suppose z ∈ K. Then either z = 0, or z = p or z = p for some p > 0 and x ∈ F. If z = 0, then ψ ≥ 0. In the first remaining case x + y = y - ∈ K, and so by definition. Thus In the second case, x - y ∈ K, and so similarly by definition and so In all cases, ψ ≥ 0, and so ψ is K-positive. We now prove that -∞ < a ≤ b. Notice by assumption there exists at least one x ∈ F for which y - x ∈ K, and so -∞ <a. However, it may be the case that there are no x ∈ F for which x - y∈ K, in which case b = ∞ and the inequality is trivial. Therefore, we may assume that b < ∞ and there is at least one x ∈ F for which x - y∈ K. To prove the inequality, it suffices to show that whenever x ∈ F and y - x ∈ K, and x' ∈ F and x' - y ∈ K, then φ ≤ φ. Indeed, since K is a convex cone, and so since φ is K-positive.
Corollary: Krein's extension theorem
Let E be a reallinear space, and let K ⊂ E be a convex cone. Let x ∈ E\ be such that Rx + K = E. Then there exists a K-positive linear functional φ: E → R such that φ > 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 × 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 > N, we have: ∈ K, whereas leading to a contradiction.
