M. Riesz extension theorem

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 ⊂ 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:

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

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:

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

In general, a K-positive linear functional on F cannot be extended to a $K$-positive linear functional on E. Already in two dimensions one obtains a counterexample taking K to be the upper halfplane 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

By transfinite induction it is sufficient to consider the case dim E/F = 1.

Choose y ∈ E\F. Set

$\psi |_{F}=\phi ,\quad \psi (y)=\sup \left\{\phi (x)\,\mid \,x\in F,\,y-x\in K\right\},$

and extend ψ to E by linearity. Let us show that ψ is K-positive.

Every point z in K is a positive linear multiple of either x + y or x − y for some x ∈ F. In the first case, z = a(y + x), therefore y− (−x) = z/a  is in  K  with  −x  in  F . Hence

$\psi (y)\geq \psi (-x)=-\psi (x),$

therefore ψ(z) ≥ 0. In the second case, z = a(x − y), therefore y = x − z/a. Let x₁ ∈ F be such that z₁ = y − x₁ ∈ K and ψ(x₁) ≥ ψ(y) − ε. Then

$\psi (x)-\psi (x_{1})=\psi (x-x_{1})=\psi (z_{1}+z/a)=\phi (z_{1}+z/a)\geq 0~,$

therefore ψ(z) ≥ −a ε. Since this is true for arbitrary ε > 0, we obtain ψ(z) ≥ 0.

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:

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

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

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

Define a functional φ₁ on R×U by

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

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

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

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

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

leading to a contradiction.

References

• Castillo, Reńe E. (2005), "A note on Krein's theorem" (PDF), Lecturas Matematicas, 26
• 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