TheM. Riesz extension theorem is atheorem inmathematics, proved byMarcel Riesz^[1] during his study of theproblem of moments.^[2]

Let${\displaystyle E}$  be arealvector space,${\displaystyle F\subset E}$  be avector subspace, and${\displaystyle K\subset E}$  be aconvex cone.

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

${\displaystyle \varphi (x)\ge 0\text{for}x\in F\cap K.}$ 

A linear functional${\displaystyle \psi :E\to \mathbb{R}}$  is called a${\displaystyle K}$ -positiveextension of${\displaystyle \varphi }$ , if it is identical to${\displaystyle \varphi }$  in the domain of${\displaystyle \varphi }$ , and also returns a value of at least 0 for all points in the cone${\displaystyle K}$ :

${\displaystyle \psi {|}_F=\varphi \text{and}\psi (x)\ge 0\text{for}x\in K.}$ 

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

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

The proof is similar to the proof of theHahn–Banach theorem (see also below).

Bytransfinite induction orZorn's lemma it is sufficient to consider the case dim ${\displaystyle E/F=1}$ .

Choose any${\displaystyle y\in E\setminus F}$ . Set

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

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

${\displaystyle \psi (y)=c\ge a\ge \varphi (-x)=\psi (-x)}$ 

by definition. Thus

${\displaystyle \psi (z)=p\psi (x+y)=p(\psi (x)+\psi (y))\ge 0.}$ 

In the second case,${\displaystyle x-y\in K}$ , and so similarly

${\displaystyle \psi (y)=c\le b\le \varphi (x)=\psi (x)}$ 

by definition and so

${\displaystyle \psi (z)=p\psi (x-y)=p(\psi (x)-\psi (y))\ge 0.}$ 

In all cases,${\displaystyle \psi (z)>0}$ , and so${\displaystyle \psi }$  is${\displaystyle K}$ -positive.

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

${\displaystyle x^{\prime }-x=(x^{\prime }-y)+(y-x)\in K}$ 

since${\displaystyle K}$  is a convex cone, and so

${\displaystyle 0\le \varphi (x^{\prime }-x)=\varphi (x^{\prime })-\varphi (x)}$ 

since${\displaystyle \varphi }$  is${\displaystyle K}$ -positive.

LetE be areallinear space, and letK ⊂ E be aconvex cone. Letx ∈ E/(−K) be such thatR x + K = E. Then there exists aK-positive linear functionalφ: E → R such thatφ(x) > 0.

The Hahn–Banach theorem can be deduced from the M. Riesz extension theorem.

LetV be a linear space, and letN be a sublinear function onV. Letφ be a functional on a subspaceU ⊂ V that is dominated byN:

${\displaystyle \varphi (x)\le N(x),x\in U.}$ 

The Hahn–Banach theorem asserts thatφ can be extended to a linear functional onV that is dominated byN.

To derive this from the M. Riesz extension theorem, define a convex coneK ⊂ R×V by

${\displaystyle K=\left\{(a,x)\mid N(x)\le a\right\}.}$ 

Define a functionalφ₁ onR×U by

${\displaystyle {\varphi }_1(a,x)=a-\varphi (x).}$ 

One can see thatφ₁ isK-positive, and thatK + (R × U) = R × V. Thereforeφ₁ can be extended to aK-positive functionalψ₁ onR×V. Then

${\displaystyle \psi (x)=-{\psi }_1(0,x)}$ 

is the desired extension ofφ. Indeed, ifψ(x) > N(x), we have: (N(x), x) ∈ K, whereas

 

leading to a contradiction.

• Castillo, Reńe E. (2005),"A note on Krein's theorem"(PDF),Lecturas Matematicas,26, archived fromthe original(PDF) on 2014-02-01, retrieved2014-01-18
• Riesz, M. (1923), "Sur le problème des moments. III.",Arkiv för Matematik, Astronomi och Fysik (in French),17 (16),JFM 49.0195.01
• Akhiezer, N.I. (1965),The classical moment problem and some related questions in analysis, New York: Hafner Publishing Co.,MR 0184042