# 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]

## 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 $\displaystyle{ E=\mathbb{R}^2,\ K=\{(x,y): y\gt 0\}\cup\{(x,0): x\gt 0\}, }$ 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 $\displaystyle{ -\infty \lt a \le b }$. For now, choose any $\displaystyle{ c }$ satisfying $\displaystyle{ a \le c \le b }$, 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 $\displaystyle{ p \gt 0 }$ and $\displaystyle{ x \in F }$. If $\displaystyle{ z = 0 }$, then $\displaystyle{ \psi(z) \gt 0 }$. 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, $\displaystyle{ \psi(z) \gt 0 }$, and so $\displaystyle{ \psi }$ is $\displaystyle{ K }$-positive.

We now prove that $\displaystyle{ -\infty \lt a \le b }$. Notice by assumption there exists at least one $\displaystyle{ x \in F }$ for which $\displaystyle{ y - x \in K }$, and so $\displaystyle{ -\infty \lt a }$. 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 $\displaystyle{ b \lt \infty }$ 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 $\displaystyle{ \phi(x) \le \phi(x') }$. 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 φE → 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:

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

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

$\displaystyle{ \psi_1(N(x), x) = N(x) - \psi(x) \lt 0, }$