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 {\displaystyle E} be a real vector space, F ⊂ E {\displaystyle F\subset E} be a vector subspace, and K ⊂ E {\displaystyle K\subset E} be a convex cone. A linear functional ϕ : F → R {\displaystyle \phi :F\to \mathbb {R} } is called K {\displaystyle K} -positive, if it takes only non-negative values on the cone K {\displaystyle K} :
ϕ ( x ) ≥ 0 for x ∈ F ∩ K . {\displaystyle \phi (x)\geq 0\quad {\text{for}}\quad x\in F\cap K.}
A linear functional ψ : E → R {\displaystyle \psi :E\to \mathbb {R} } is called a K {\displaystyle K} -positive extension of ϕ {\displaystyle \phi } , if it is identical to ϕ {\displaystyle \phi } in the domain of ϕ {\displaystyle \phi } , and also returns a value of at least 0 for all points in the cone K {\displaystyle K} :
ψ | F = ϕ and ψ ( x ) ≥ 0 for x ∈ K . {\displaystyle \psi |_{F}=\phi \quad {\text{and}}\quad \psi (x)\geq 0\quad {\text{for}}\quad x\in K.}
In general, a K {\displaystyle K} -positive linear functional on F {\displaystyle F} cannot be extended to a K {\displaystyle K} -positive linear functional on E {\displaystyle E} . Already in two dimensions one obtains a counterexample. Let E = R 2 , K = { ( x , y ) : y > 0 } ∪ { ( x , 0 ) : x > 0 } , {\displaystyle E=\mathbb {R} ^{2},\ K=\{(x,y):y>0\}\cup \{(x,0):x>0\},} and F {\displaystyle F} be the x {\displaystyle x} -axis. The positive functional ϕ ( x , 0 ) = x {\displaystyle \phi (x,0)=x} can not be extended to a positive functional on E {\displaystyle E} . However, the extension exists under the additional assumption that E ⊂ K + F , {\displaystyle E\subset K+F,} namely for every y ∈ E , {\displaystyle y\in E,} there exists an x ∈ F {\displaystyle x\in F} such that y − x ∈ K . {\displaystyle 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 {\displaystyle E/F=1} . Choose any y ∈ E ∖ F {\displaystyle y\in E\setminus F} . Set
a = sup { ϕ ( x ) ∣ x ∈ F , y − x ∈ K } , b = inf { ϕ ( x ) ∣ x ∈ F , x − y ∈ K } . {\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\,\}.}
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