A linear functional is called -positive, if it takes only non-negative values on the cone :
A linear functional is called a -positive extension of , if it is identical to in the domain of , and also returns a value of at least 0 for all points in the cone :
In general, a -positive linear functional on cannot be extended to a -positive linear functional on . Already in two dimensions one obtains a counterexample. Let and be the -axis. The positive functional can not be extended to a positive functional on .
However, the extension exists under the additional assumption that namely for every there exists an such that
Proofedit
The proof is similar to the proof of the Hahn–Banach theorem (see also below).
We will prove below that . For now, choose any satisfying , and set , , and then extend to all of by linearity. We need to show that is -positive. Suppose . Then either , or or for some and . If , then . In the first remaining case , and so
by definition. Thus
In the second case, , and so similarly
by definition and so
In all cases, , and so is -positive.
We now prove that . Notice by assumption there exists at least one for which , and so . However, it may be the case that there are no for which , in which case 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 for which . To prove the inequality, it suffices to show that whenever and , and and , then . Indeed,
since is a convex cone, and so
since is -positive.
Corollary: Krein's extension theoremedit
Let E be a reallinear space, and let K ⊂ E be a convex cone. Let x ∈ E/(−K) be such that Rx + K = E. Then there exists a K-positive linear functional φ: E → R such that φ(x) > 0.
Castillo, Reńe E. (2005), (PDF), Lecturas Matematicas, 26, archived from the original (PDF) on 2014-02-01, retrieved 2014-01-18
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
March 21, 2024
riesz, extension, theorem, more, theorems, that, sometimes, called, riesz, theorem, riesz, theorem, theorem, mathematics, proved, marcel, riesz, during, study, problem, moments, contents, formulation, proof, corollary, krein, extension, theorem, connection, ha. For more theorems that are sometimes called Riesz s theorem see Riesz theorem The M Riesz extension theorem is a theorem in mathematics proved by Marcel Riesz 1 during his study of the problem of moments 2 Contents 1 Formulation 2 Proof 3 Corollary Krein s extension theorem 4 Connection to the Hahn Banach theorem 5 References 6 SourcesFormulation editLet E displaystyle E nbsp be a real vector space F E displaystyle F subset E nbsp be a vector subspace and K E displaystyle K subset E nbsp be a convex cone A linear functional ϕ F R displaystyle phi F to mathbb R nbsp is called K displaystyle K nbsp positive if it takes only non negative values on the cone K displaystyle K nbsp ϕ x 0 for x F K displaystyle phi x geq 0 quad text for quad x in F cap K nbsp A linear functional ps E R displaystyle psi E to mathbb R nbsp is called a K displaystyle K nbsp positive extension of ϕ displaystyle phi nbsp if it is identical to ϕ displaystyle phi nbsp in the domain of ϕ displaystyle phi nbsp and also returns a value of at least 0 for all points in the cone K displaystyle K nbsp ps F ϕ and ps x 0 for x K displaystyle psi F phi quad text and quad psi x geq 0 quad text for quad x in K nbsp In general a K displaystyle K nbsp positive linear functional on F displaystyle F nbsp cannot be extended to a K displaystyle K nbsp positive linear functional on E displaystyle E nbsp Already in two dimensions one obtains a counterexample Let E R 2 K x y y gt 0 x 0 x gt 0 displaystyle E mathbb R 2 K x y y gt 0 cup x 0 x gt 0 nbsp and F displaystyle F nbsp be the x displaystyle x nbsp axis The positive functional ϕ x 0 x displaystyle phi x 0 x nbsp can not be extended to a positive functional on E displaystyle E nbsp However the extension exists under the additional assumption that E K F displaystyle E subset K F nbsp namely for every y E displaystyle y in E nbsp there exists an x F displaystyle x in F nbsp such that y x K displaystyle y x in K nbsp Proof editThe 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 nbsp Choose any y E F displaystyle y in E setminus F nbsp 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 nbsp We will prove below that lt a b displaystyle infty lt a leq b nbsp For now choose any c displaystyle c nbsp satisfying a c b displaystyle a leq c leq b nbsp and set ps y c displaystyle psi y c nbsp ps F ϕ displaystyle psi F phi nbsp and then extend ps displaystyle psi nbsp to all of E displaystyle E nbsp by linearity We need to show that ps displaystyle psi nbsp is K displaystyle K nbsp positive Suppose z K displaystyle z in K nbsp Then either z 0 displaystyle z 0 nbsp or z p x y displaystyle z p x y nbsp or z p x y displaystyle z p x y nbsp for some p gt 0 displaystyle p gt 0 nbsp and x F displaystyle x in F nbsp If z 0 displaystyle z 0 nbsp then ps z gt 0 displaystyle psi z gt 0 nbsp In the first remaining case x y y x K displaystyle x y y x in K nbsp and so ps y c a ϕ x ps x displaystyle psi y c geq a geq phi x psi x nbsp by definition Thus ps z p ps x y p ps x ps y 0 displaystyle psi z p psi x y p psi x psi y geq 0 nbsp In the second case x y K displaystyle x y in K nbsp and so similarly ps y c b ϕ x ps x displaystyle psi y c leq b leq phi x psi x nbsp by definition and so ps z p ps x y p ps x ps y 0 displaystyle psi z p psi x y p psi x psi y geq 0 nbsp In all cases ps z gt 0 displaystyle psi z gt 0 nbsp and so ps displaystyle psi nbsp is K displaystyle K nbsp positive We now prove that lt a b displaystyle infty lt a leq b nbsp Notice by assumption there exists at least one x F displaystyle x in F nbsp for which y x K displaystyle y x in K nbsp and so lt a displaystyle infty lt a nbsp However it may be the case that there are no x F displaystyle x in F nbsp for which x y K displaystyle x y in K nbsp in which case b displaystyle b infty nbsp and the inequality is trivial in this case notice that the third case above cannot happen Therefore we may assume that b lt displaystyle b lt infty nbsp and there is at least one x F displaystyle x in F nbsp for which x y K displaystyle x y in K nbsp To prove the inequality it suffices to show that whenever x F displaystyle x in F nbsp and y x K displaystyle y x in K nbsp and x F displaystyle x in F nbsp and x y K displaystyle x y in K nbsp then ϕ x ϕ x displaystyle phi x leq phi x nbsp Indeed x x x y y x K displaystyle x x x y y x in K nbsp since K displaystyle K nbsp is a convex cone and so 0 ϕ x x ϕ x ϕ x displaystyle 0 leq phi x x phi x phi x nbsp since ϕ displaystyle phi nbsp is K displaystyle K nbsp positive Corollary Krein s extension theorem editLet 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 f E R such that f x gt 0 Connection to the Hahn Banach theorem editMain article 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 f be a functional on a subspace U V that is dominated by N ϕ x N x x U displaystyle phi x leq N x quad x in U nbsp The Hahn Banach theorem asserts that f 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 a x N x a displaystyle K left a x mid N x leq a right nbsp Define a functional f1 on R U by ϕ 1 a x a ϕ x displaystyle phi 1 a x a phi x nbsp One can see that f1 is K positive and that K R U R V Therefore f1 can be extended to a K positive functional ps1 on R V Then ps x ps 1 0 x displaystyle psi x psi 1 0 x nbsp is the desired extension of f Indeed if ps x gt N x we have N x x K whereas ps 1 N x x N x ps x lt 0 displaystyle psi 1 N x x N x psi x lt 0 nbsp leading to a contradiction References edit Riesz 1923 Akhiezer 1965 Sources editCastillo Rene E 2005 A note on Krein s theorem PDF Lecturas Matematicas 26 archived from the original PDF on 2014 02 01 retrieved 2014 01 18 Riesz M 1923 Sur le probleme des moments III Arkiv for 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 Retrieved from https en wikipedia org w index php title M Riesz extension theorem amp oldid 1211051283, wikipedia, wiki, book, books, library,