M. Riesz extension theorem explained

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\subset E be a vector subspace, and K\subset E be a convex cone.\phi: F\to R is called K -positive, if it takes only non-negative values on the cone K :\phi(x) \geq 0 for x \in F \cap K.

A linear functional

\psi: E\to R is called a K -positive extension of \phi , if it is identical to \phi in the domain of \phi , and also returns a value of at least 0 for all points in the cone K :\psi| _{F} = \phi and \psi(x) \geq 0 for 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. Let E = R ^{2, K=\{(x,y):} y> 0 \}\cup\{(x, 0 ): x> 0 \}, and F be the x -axis. The positive functional \phi(x, 0 ) = x can not be extended to a positive functional on E .

However, the extension exists under the additional assumption that

E\subset K + F, namely for every y\in E, there exists an x\in F such that y - x\in K.

Proof

The proof is similar to the proof of the Hahn–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 \in E \setminus F . Seta = \sup \{ \phi(x) \mid x \in F, y - x \in K \}, b = inf \{ \phi(x) \mid x \in F, x - y \in K \}.

We will prove below that

-inf ty < a \le b . For now, choose any c satisfying a \le c \le b , and set \psi(y) = c , \psi| _{F} = \phi , and then extend \psi to all of E by linearity. We need to show that \psi is K -positive. Suppose z \in K . Then either z = 0 , or z = p(x + y) or z = p(x - y) for some p > 0 and x \in F . If z = 0 , then \psi(z) > 0 . In the first remaining case x + y = y - ( - x) \in K , and so\psi(y) = c \geq a \geq \phi( - x) = \psi( - x)

by definition. Thus

\psi(z) = p\psi(x + y) = p(\psi(x) + \psi(y)) \geq 0 .

In the second case,

x - y \in K , and so similarly\psi(y) = c \leq b \leq \phi(x) = \psi(x)

by definition and so

\psi(z) = p\psi(x - y) = p(\psi(x) - \psi(y)) \geq 0 .

In all cases,

\psi(z) > 0 , and so \psi is K -positive.

We now prove that

-inf ty < a \le b . Notice by assumption there exists at least one x \in F for which y - x \in K , and so -inf ty < a . However, it may be the case that there are no x \in F for which x - y \in K , in which case b = inf ty and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that b < inf ty and there is at least one x \in F for which x - y \in K . To prove the inequality, it suffices to show that whenever x \in F and y - x \in K , and x' \in F and x' - y \in K , then \phi(x) \le \phi(x') . Indeed,x' - x = (x' - y) + (y - x) \in K

since

K is a convex cone, and so0 \leq \phi(x' - x) = \phi(x') - \phi(x)

since

\phi 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

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), 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.