# 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 $\phi :F\to \mathbb {R}$ is called $K$-positive, if it takes only non-negative values on the cone $K$:
    $\phi (x)\geq 0\quad {\text{for}}\quad x\in F\cap K.$
A linear functional $\psi :E\to \mathbb {R}$ is called a $K$-positive extension of $\phi$, if it is identical to $\phi$ in the domain of $\phi$, and also returns a value of at least 0 for all points in the cone $K$:
    $\psi |_{F}=\phi \quad {\text{and}}\quad \psi (x)\geq 0\quad {\text{for}}\quad x\in K.$
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 $E=\mathbb {R} ^{2},\ K=\{(x,y):y>0\}\cup \{(x,0):x>0\},$ and $F$ be the $x$-axis. The positive functional $\phi (x,0)=x$ 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
    $a=\sup\{\,\phi (x)\mid x\in F,\ y-x\in K\,\},\ b=\inf\{\,\phi (x)\mid x\in F,x-y\in K\,\}.$
We will prove below that $-\infty <a\leq b$. For now, choose any $c$ satisfying $a\leq c\leq b$, and set $\psi (y)=c$, $\psi |_{F}=\phi$, 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
    $\psi (y)=c\geq a\geq \phi (-x)=\psi (-x)$
by definition. Thus
    $\psi (z)=p\psi (x+y)=p(\psi (x)+\psi (y))\geq 0.$
In the second case, $x-y\in K$, and so similarly
    $\psi (y)=c\leq b\leq \phi (x)=\psi (x)$
by definition and so
    $\psi (z)=p\psi (x-y)=p(\psi (x)-\psi (y))\geq 0.$
In all cases, $\psi (z)>0$, and so $\psi$ is $K$-positive.
We now prove that $-\infty <a\leq b$. Notice by assumption there exists at least one $x\in F$ for which $y-x\in K$, and so $-\infty <a$. However, it may be the case that there are no $x\in F$ for which $x-y\in K$, in which case $b=\infty$ and the inequality is trivial (in this case notice that the third case above cannot happen). Therefore, we may assume that $b<\infty$ 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 $x'\in F$ and $x'-y\in K$, then $\phi (x)\leq \phi (x')$. Indeed,
    $x'-x=(x'-y)+(y-x)\in K$
since $K$ is a convex cone, and so
    $0\leq \phi (x'-x)=\phi (x')-\phi (x)$
since $\phi$ 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:
    $\phi (x)\leq N(x),\quad x\in U.$
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
    $K=\left\{(a,x)\,\mid \,N(x)\leq a\right\}.$
Define a functional φ$_1$ on R×U by
    $\phi _{1}(a,x)=a-\phi (x).$
One can see that φ$_1$ is K-positive, and that K + (R × U) = R × V. Therefore φ$_1$ can be extended to a K-positive functional ψ$_1$ 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.