# 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 φF → R is called K-positive, if

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

A linear functional ψE → R is called a K-positive extension of φ if

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

In general, a K-positive linear functional on F can not 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 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 xF 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 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:

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

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

$\psi_1(N(x), x) = N(x) - \psi(x) < 0,$