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 $\phi: F\to\mathbb$ is called $K$-''positive'', if it takes only non-negative values on the cone $K$:

:$\phi(x) \geq 0 \quad \text \quad x \in F \cap K.$

A linear functional $\psi: E\to\mathbb$ 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 $K$:

:$\psi, _F = \phi \quad \text \quad \psi(x) \geq 0\quad \text \quad x \in K.$

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 $E=\mathbb^2,\ K=\\cup\,$ 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

:$a = \sup \,\ b = \inf \.$

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'' ⊂ ''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

:$K = \left\.$

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.

Notes

References

*
*
*

{{Functional Analysis

Theorems in convex geometry
Theorems in functional analysis