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M. Riesz extension theorem

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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 φF → R is called K-positive, if it takes only non-negative values on the cone K:

A linear functional ψE → 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 xF 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 yE\F. Set

We will prove below that -∞ < ab. For now, choose any c satisfying acb, 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 zK. Then either z = 0, or z = p(x + y) or z = p(x - y) for some p > 0 and xF. 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 - yK, and so similarly

by definition and so

In all cases, ψ(z) ≥ 0, and so ψ is K-positive.

We now prove that -∞ < ab. Notice by assumption there exists at least one xF for which y - xK, and so -∞ <a. However, it may be the case that there are no x ∈ F for which x - yK, 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 - yK. To prove the inequality, it suffices to show that whenever xF and y - xK, and x'F and x' - yK, 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 φE → 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