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 $$-\infty < a \le b$$. For now, choose any $$c$$ satisfying $$a \le c \le b$$, 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 $$-\infty < a \le b$$. 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 $$\phi(x) \le \phi(x')$$. 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 &sub; E be a convex cone. Let x &isin; E/(&minus;K) be such that R x + K = E. Then there exists a K-positive linear functional &phi;: E &rarr; R such that &phi;(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 &phi; be a functional on a subspace U &sub; V that is dominated by N:


 * $$ \phi(x) \leq N(x), \quad x \in U.$$

The Hahn–Banach theorem asserts that &phi; 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 &sub; R&times;V by


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

Define a functional &phi;1 on R&times;U by


 * $$ \phi_1(a, x) = a - \phi(x).$$

One can see that &phi;1 is K-positive, and that K + (R &times; U) = R &times; V. Therefore &phi;1 can be extended to a K-positive functional &psi;1 on R&times;V. Then


 * $$ \psi(x) = - \psi_1(0, x) $$

is the desired extension of &phi;. Indeed, if &psi;(x) > N(x), we have: (N(x), x) &isin; K, whereas


 * $$ \psi_1(N(x), x) = N(x) - \psi(x) < 0, $$

leading to a contradiction.