M. Riesz extension theorem
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 can not be extended to a -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 real 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.
By transfinite induction it is sufficient to consider the case dim E/F = 1.
Choose y ∈ E\F. Set
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
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
therefore ψ(z) ≥ −a ε. Since this is true for arbitrary ε > 0, we obtain ψ(z) ≥ 0.
Corollary: Krein's extension theorem
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.
- Castillo, Reńe E. (2005), "A note on Krein’s theorem", Lecturas Matematicas 26
- Riesz, M. (1923), "Sur le problème des moments. III.", Ark. F. Mat. Astr. O. Fys. (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