M. Riesz extension 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]

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

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

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

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

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

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

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

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 ${\displaystyle E/F=1}${\displaystyle E/F=1}.

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

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

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

${\displaystyle \psi (y)=c\geq a\geq \phi (-x)=\psi (-x)}${\displaystyle \psi (y)=c\geq a\geq \phi (-x)=\psi (-x)}

by definition. Thus

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

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

${\displaystyle \psi (y)=c\leq b\leq \phi (x)=\psi (x)}${\displaystyle \psi (y)=c\leq b\leq \phi (x)=\psi (x)}

by definition and so

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

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

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

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

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

${\displaystyle 0\leq \phi (x'-x)=\phi (x')-\phi (x)}${\displaystyle 0\leq \phi (x'-x)=\phi (x')-\phi (x)}

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

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.

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:

${\displaystyle \phi (x)\leq N(x),\quad x\in U.}${\displaystyle \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

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

Define a functional φ₁ on R×U by

${\displaystyle \phi _{1}(a,x)=a-\phi (x).}${\displaystyle \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

${\displaystyle \psi (x)=-\psi _{1}(0,x)}${\displaystyle \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

${\displaystyle \psi _{1}(N(x),x)=N(x)-\psi (x)<0,}${\displaystyle \psi _{1}(N(x),x)=N(x)-\psi (x)<0,}

leading to a contradiction.

• Castillo, Reńe E. (2005), "A note on Krein's theorem" (PDF), Lecturas Matematicas, 26, archived from the original (PDF) on 2014年02月01日, retrieved 2014年01月18日
• 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

• Theorems in convex geometry
• Theorems in functional analysis

• CS1 French-language sources (fr)