Extension of positive functionals II

Question

This is a follow-up to

Extension of positive functionals.

Assume that $X=R^n$ with the canonical order (I indicate with $K$ the positive cone, $x \in K$ iff $x_i \ge 0$ for all $i$) and let $L:M \to R$ is a positive functional defined on a subspace of $X$. Then it should be true that $L$ has a positive extension to $X$. Actually, there is an old post here

https://math.stackexchange.com/questions/18593/extending-a-positive-linear-functional-in-finite-dimensions

where the author asks for a simpler proof of the result, but the original proof is not shown.

If $M$ contains an interior point of $K$ the extension follows from Krein-Rutman and if $M \cap K=\{0\}$ (or even if $(Ker L) \cap K=\{0\}$) from Bauer-Namioka. What to do with in the remaining cases? One could add $\epsilon$-times a strictly positive functional but I do not know how to control the norm of the extensions to let $\epsilon \to 0$.

So the question is about a proof of the above result, if this is true as it should be. Finally, I do not know what happens for more general (closed) cones in $R^n$. Thank you to all

Answer 1

I answer myself but maybe somebody is interested and I give a sketch of the proof based on more general ideas by Mirkil and Klee. The main point is the following

Lemma 1. Let $f_1, \cdots, f_k$ be vectors in $X=R^n$ and $G=\{\sum_{i=1}^k a_if_i, \ a_i \geq 0 \}$. Then $G$ is closed.

This is obvious if the vectors are linearly independent. The proof in the general case is by induction on $k$ and holds in any Banach space.

The following is an easy consequence

Lemma 2. If $M$ is a subspace of $X$, then $M+K$ is closed.

For the proof one considers the quotient map $q$ from $X \to X/M$ (or the orthogonal projection onto $M^{\perp}$) and applies Lemma 1 in $X/M$ to $q(K)$ with $f_i=q(e_i)$ (here $(e_i)$ is the standard basis of $X$). Then $q(K)$ closed in $X/M$ gives $M+K=q^{-1}(q(K))$ closed in $X$.

The proof of the extension of any positive map $f:M \to R$ goes as follows.

Write $f(x)=(x, x_0)$ for some $x_0 \in M$. We look for $\bar x \in K$ such that $\bar x-x_0 \in M^{\perp}$, so that $x \to (\bar x, x)$ is a positive extension. If such a $\bar x$ does not exist, then $x_0 \not \in K+M^\perp$ which is closed. Hahn-Banach then gives $y$ such that $(x_0,y) <(k+g,y)$ for every $k \in K, g\in M^\perp$. If $k=0$, then $(x_0,y) < (g,y)$ for any $g \in M^\perp$ yields $y \in M$. But then $(x_0,y)<(k,y)$ for every $k \in K$ gives $(k,y) \ge 0$ for every $k \in K$, hence $y \in K$. Finally, $y \in M\cap K$ but $(x_0,y)<(0,y)=0$, against the positivity of $f$.

Two remarks

1. The same proof works for polyedral cones even in infinite dimension. One has to work with the dual cone in $X'$.

2. The condition $d(x,M \cap(-K)) \le Cd(x,-K)$, see the comment by @fedja, would give a positive extension as an application of the Hahn-Banach theorem. However, I cannot verify it for a general subspace $M$ and I do not know if this is necessary. Please, let know.