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:

${\displaystyle \phi (x)\geq 0\quad {\text{for}}\quad x\in F\cap K.}$

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:

${\displaystyle \psi |_{F}=\phi \quad {\text{and}}\quad \psi (x)\geq 0\quad {\text{for}}\quad x\in K.}$

In general, a K-positive linear functional on F cannot be extended to a ${\displaystyle K}$-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

${\displaystyle a=\sup\{\phi (x):x\in F,\ y-x\in K\},\ b=\inf\{\phi (x):x\in F,x-y\in K\}.}$

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

${\displaystyle \psi (y)=c\geq a\geq \phi (-x)=\psi (-x)}$

by definition. Thus

${\displaystyle \psi (z)=p\psi (x+y)=p(\psi (x)+\psi (y))\geq 0.}$

In the second case, x - y ∈ K, and so similarly

${\displaystyle \psi (y)=c\leq b\leq \phi (x)=\psi (x)}$

by definition and so

${\displaystyle \psi (z)=p\psi (x-y)=p(\psi (x)-\psi (y))\geq 0.}$

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,

${\displaystyle x'-x=(x'-y)+(y-x)\in K}$

since K is a convex cone, and so

${\displaystyle 0\leq \phi (x'-x)=\phi (x')-\phi (x)}$

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

Main article: 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:

${\displaystyle \phi (x)\leq N(x),\quad x\in U.}$

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

${\displaystyle K=\left\{(a,x)\,\mid \,N(x)\leq a\right\}.}$

Define a functional φ₁ on R×U by

${\displaystyle \phi _{1}(a,x)=a-\phi (x).}$

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

${\displaystyle \psi (x)=-\psi _{1}(0,x)}$

is the desired extension of φ. Indeed, if ψ(x) > N(x), we have: (N(x), x) ∈ K, whereas

${\displaystyle \psi _{1}(N(x),x)=N(x)-\psi (x)<0,}$

leading to a contradiction.

Notes

1. Riesz (1923)
2. Akhiezer (1965)

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