# Punchscan

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 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

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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

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