M. Riesz extension theorem#Corollary: Krein's extension theorem

{{About||more theorems that are sometimes called Riesz's theorem|Riesz theorem (disambiguation){{!}}Riesz theorem}}

The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz{{harvtxt|Riesz|1923}} during his study of the problem of moments.{{harvtxt|Akhiezer|1965}}

Formulation

Let $E$ be a real vector space, $F\subset E$ be a vector subspace, and $K\subset E$ be a convex cone.

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

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

A linear functional $\psi: E\to\mathbb{R}$ 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 $K$ :

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

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 $E=\mathbb{R}^2,\ K=\{(x,y): y>0\}\cup\{(x,0): x>0\},$ 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 $y-x\in K.$

{{Math theorem

| name = {{visible anchor|M. Riesz extension theorem}}

| math_statement = Let $E$ be a real vector space, $F\subset E$ a subspace, and $K\subset E$ a convex cone verifying $E\subset K+F$ . Then every continuous, $K$ -positive, linear functional $\phi\colon F\to\mathbb R$ has a $K$ -positive extension to all of $E$ .

}}

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 $y \in E \setminus F$ . Set

: $a = \sup \{\, \phi(x) \mid x \in F, \ y-x \in K \,\},\ b = \inf \{\, \phi(x) \mid x \in F, x-y \in K \,\}.$

We will prove below that $-\infty < a \le b$ . 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 $-\infty < a \le b$ . Notice by assumption there exists at least one $x \in F$ for which $y - x \in K$ , and so $-\infty < a$ . However, it may be the case that there are no $x \in F$ for which $x - y \in K$ , in which case $b = \infty$ and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that $b < \infty$ 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 so

: $0 \leq \phi(x'-x) = \phi(x')-\phi(x)$

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:

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

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

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

leading to a contradiction.

References

Sources

{{Citation|last=Castillo|first=Reńe E.|title=A note on Krein's theorem|url=http://www.scm.org.co/aplicaciones/revista/Articulos/767.pdf|journal=Lecturas Matematicas|volume=26|year=2005|access-date=2014-01-18|archive-url=https://web.archive.org/web/20140201131652/http://www.scm.org.co/aplicaciones/revista/Articulos/767.pdf|archive-date=2014-02-01|url-status=dead}}
{{Citation|jfm=49.0195.01|last=Riesz|first=M.|title=Sur le problème des moments. III.|language=French|journal=Arkiv för Matematik, Astronomi och Fysik |volume=17|issue=16|year=1923}}
{{Citation|mr=0184042|first=N.I.|last=Akhiezer|title=The classical moment problem and some related questions in analysis|publisher=Hafner Publishing Co.|location=New York|year=1965}}

Category:Theorems in convex geometry

Category:Theorems in functional analysis