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 [math]\displaystyle{ E }[/math] be a real vector space, [math]\displaystyle{ F\subset E }[/math] be a vector subspace, and [math]\displaystyle{ K\subset E }[/math] be a convex cone.

A linear functional [math]\displaystyle{ \phi: F\to\mathbb{R} }[/math] is called [math]\displaystyle{ K }[/math]-positive, if it takes only non-negative values on the cone [math]\displaystyle{ K }[/math]:

[math]\displaystyle{ \phi(x) \geq 0 \quad \text{for} \quad x \in F \cap K. }[/math]

A linear functional [math]\displaystyle{ \psi: E\to\mathbb{R} }[/math] is called a [math]\displaystyle{ K }[/math]-positive extension of [math]\displaystyle{ \phi }[/math], if it is identical to [math]\displaystyle{ \phi }[/math] in the domain of [math]\displaystyle{ \phi }[/math], and also returns a value of at least 0 for all points in the cone [math]\displaystyle{ K }[/math]:

[math]\displaystyle{ \psi|_F = \phi \quad \text{and} \quad \psi(x) \geq 0\quad \text{for} \quad x \in K. }[/math]

In general, a [math]\displaystyle{ K }[/math]-positive linear functional on [math]\displaystyle{ F }[/math] cannot be extended to a [math]\displaystyle{ K }[/math]-positive linear functional on [math]\displaystyle{ E }[/math]. Already in two dimensions one obtains a counterexample. Let [math]\displaystyle{ E=\mathbb{R}^2,\ K=\{(x,y): y\gt 0\}\cup\{(x,0): x\gt 0\}, }[/math] and [math]\displaystyle{ F }[/math] be the [math]\displaystyle{ x }[/math]-axis. The positive functional [math]\displaystyle{ \phi(x,0)=x }[/math] can not be extended to a positive functional on [math]\displaystyle{ E }[/math].

However, the extension exists under the additional assumption that [math]\displaystyle{ E\subset K+F, }[/math] namely for every [math]\displaystyle{ y\in E, }[/math] there exists an [math]\displaystyle{ x\in F }[/math] such that [math]\displaystyle{ y-x\in K. }[/math]

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 [math]\displaystyle{ E/F = 1 }[/math].

Choose any [math]\displaystyle{ y \in E \setminus F }[/math]. Set

[math]\displaystyle{ a = \sup \{\, \phi(x) \mid x \in F, \ y-x \in K \,\},\ b = \inf \{\, \phi(x) \mid x \in F, x-y \in K \,\}. }[/math]

We will prove below that [math]\displaystyle{ -\infty \lt a \le b }[/math]. For now, choose any [math]\displaystyle{ c }[/math] satisfying [math]\displaystyle{ a \le c \le b }[/math], and set [math]\displaystyle{ \psi(y) = c }[/math], [math]\displaystyle{ \psi|_F = \phi }[/math], and then extend [math]\displaystyle{ \psi }[/math] to all of [math]\displaystyle{ E }[/math] by linearity. We need to show that [math]\displaystyle{ \psi }[/math] is [math]\displaystyle{ K }[/math]-positive. Suppose [math]\displaystyle{ z \in K }[/math]. Then either [math]\displaystyle{ z = 0 }[/math], or [math]\displaystyle{ z = p(x + y) }[/math] or [math]\displaystyle{ z = p(x - y) }[/math] for some [math]\displaystyle{ p \gt 0 }[/math] and [math]\displaystyle{ x \in F }[/math]. If [math]\displaystyle{ z = 0 }[/math], then [math]\displaystyle{ \psi(z) \gt 0 }[/math]. In the first remaining case [math]\displaystyle{ x + y = y -(-x) \in K }[/math], and so

[math]\displaystyle{ \psi(y) = c \geq a \geq \phi(-x) = \psi(-x) }[/math]

by definition. Thus

[math]\displaystyle{ \psi(z) = p\psi(x+y) = p(\psi(x) + \psi(y)) \geq 0. }[/math]

In the second case, [math]\displaystyle{ x - y \in K }[/math], and so similarly

[math]\displaystyle{ \psi(y) = c \leq b \leq \phi(x) = \psi(x) }[/math]

by definition and so

[math]\displaystyle{ \psi(z) = p\psi(x-y) = p(\psi(x)-\psi(y)) \geq 0. }[/math]

In all cases, [math]\displaystyle{ \psi(z) \gt 0 }[/math], and so [math]\displaystyle{ \psi }[/math] is [math]\displaystyle{ K }[/math]-positive.

We now prove that [math]\displaystyle{ -\infty \lt a \le b }[/math]. Notice by assumption there exists at least one [math]\displaystyle{ x \in F }[/math] for which [math]\displaystyle{ y - x \in K }[/math], and so [math]\displaystyle{ -\infty \lt a }[/math]. However, it may be the case that there are no [math]\displaystyle{ x \in F }[/math] for which [math]\displaystyle{ x - y \in K }[/math], in which case [math]\displaystyle{ b = \infty }[/math] and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that [math]\displaystyle{ b \lt \infty }[/math] and there is at least one [math]\displaystyle{ x \in F }[/math] for which [math]\displaystyle{ x - y \in K }[/math]. To prove the inequality, it suffices to show that whenever [math]\displaystyle{ x \in F }[/math] and [math]\displaystyle{ y - x \in K }[/math], and [math]\displaystyle{ x' \in F }[/math] and [math]\displaystyle{ x' - y \in K }[/math], then [math]\displaystyle{ \phi(x) \le \phi(x') }[/math]. Indeed,

[math]\displaystyle{ x' -x = (x' - y) + (y-x) \in K }[/math]

since [math]\displaystyle{ K }[/math] is a convex cone, and so

[math]\displaystyle{ 0 \leq \phi(x'-x) = \phi(x')-\phi(x) }[/math]

since [math]\displaystyle{ \phi }[/math] is [math]\displaystyle{ K }[/math]-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:

[math]\displaystyle{ \phi(x) \leq N(x), \quad x \in U. }[/math]

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

[math]\displaystyle{ K = \left\{ (a, x) \, \mid \, N(x) \leq a \right\}. }[/math]

Define a functional φ₁ on R×U by

[math]\displaystyle{ \phi_1(a, x) = a - \phi(x). }[/math]

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

[math]\displaystyle{ \psi(x) = - \psi_1(0, x) }[/math]

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

[math]\displaystyle{ \psi_1(N(x), x) = N(x) - \psi(x) \lt 0, }[/math]

leading to a contradiction.

References

• Castillo, Reńe E. (2005), "A note on Krein's theorem", Lecturas Matematicas 26, http://www.scm.org.co/aplicaciones/revista/Articulos/767.pdf, retrieved 2014-01-18 
• Riesz, M. (1923), "Sur le problème des moments. III." (in French), Arkiv för Matematik, Astronomi och Fysik 17 (16) 
• Akhiezer, N.I. (1965), The classical moment problem and some related questions in analysis, New York: Hafner Publishing Co.