M. Riesz extension theorem

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]

Formulation

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

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

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

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

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

In general, a ${\displaystyle K}$-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 ${\displaystyle \phi (x,0)=x}$ 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 ${\displaystyle y-x\in K.}$

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 ${\displaystyle E/F=1}$.

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

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

We will prove below that ${\displaystyle -\infty <a\leq b}$. For now, choose any ${\displaystyle c}$ satisfying ${\displaystyle a\leq c\leq b}$, and set ${\displaystyle \psi (y)=c}$, ${\displaystyle \psi |_{F}=\phi }$, 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

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

by definition. Thus

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

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

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

by definition and so

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

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

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

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

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

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

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

Main 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 φ be a functional on a subspace U ⊂ V that is dominated by N:

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

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

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

Define a functional φ₁ on R×U by

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

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

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

is the desired extension of φ. Indeed, if ψ(x) > N(x), we have: (N(x), x) ∈ K, whereas

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

leading to a contradiction.

References

1. Riesz (1923)
2. Akhiezer (1965)

Sources

• Castillo, Reńe 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 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