M. Riesz extension theorem

The M. Riesz extension theorem is a theorem in mathematics, proved by Marcel Riesz during his study of the problem of moments.

Formulation

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

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

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

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

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

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

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

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 $E/F=1$.

Choose any $y\in E\setminus F$. Set

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

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

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

by definition. Thus

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

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

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

by definition and so

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

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

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

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

since $K$ is a convex cone, and so

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

since $\varphi$ is $K$-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

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:

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

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

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

Define a functional φ₁ on R×U by

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

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

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

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

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

leading to a contradiction.

Sources

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