M. Riesz extension theorem
The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz [1] during his study of the problem of moments.[2]
Formulation
Let E be a real vector space, F ⊂ E a vector subspace, and let K ⊂ E be a convex cone.
A linear functional φ: F → R is called K-positive, if it takes only non-negative values on the cone K:
A linear functional ψ: E → R is called a K-positive extension of φ, if it is identical to φ 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 functional φ(x, 0) = x can not be extended to a positive functional on the plane.
However, the extension exists under the additional assumption that for every y ∈ E there exists x∈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 (see also below).
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(y) = c, ψ|F = φ, and then extend ψ to all of E by linearity. We need to show that ψ is K-positive. Suppose z ∈ K. Then either z = 0, or z = p(x + y) or z = p(x - y) for some p > 0 and x ∈ F. If z = 0, then ψ(z) ≥ 0. In the first remaining case x + y = y - (-x) ∈ K, and so
by definition. Thus
In the second case, x - y ∈ K, and so similarly
by definition and so
In all cases, ψ(z) ≥ 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 (in this case notice that the third case above cannot happen). 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(x) ≤ φ(x'). Indeed,
since K is a convex cone, and so
since φ 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.
Notes
References
- Castillo, Reńe E. (2005), "A note on Krein's theorem" (PDF), Lecturas Matematicas, 26, archived from the original (PDF) on 2014-02-01, retrieved 2014-01-18
- Riesz, M. (1923), "Sur le problème des moments. III.", Arkiv för Matematik, Astronomi och Fysik (in French), 17 (16), JFM 49.0195.01
- Akhiezer, N.I. (1965), The classical moment problem and some related questions in analysis, New York: Hafner Publishing Co., MR 0184042