Let $(X,\leq)$ be a partially ordered real vector space, let $\Gamma\subseteq X$ be a convex cone, containing $0$, that is such that for all $x\in X$, there is $y\in\Gamma$ such that $x\leq y$ and $\forall z\in \Gamma$ such that $x\leq z$, $y\leq z$, denote such a $y$ by $\overline{x}$ and let $\underline x = -\left(\overline{-x} \right)$.
Lemma 1 : If $L:\Gamma\to\mathbb R$ is such that
- $\forall x,y\in \Gamma$, $L(x+y)=L(x)+L(y)$
- $\forall x\in\Gamma,a\in [0,\infty[$, $L(ax)=aL(x)$
- $\forall x,y\in\Gamma$ with $x\leq y$, $L(x)\leq L(y)$
Then there exists a positive linear functional $f:X\to\mathbb R$ such that for all $x\in\Gamma$, $f(x)=L(x)$. I am interested in proving the following Lemma, or adding some constraint such that it holds.
I provide a proof of this Lemma at the end of the post.
Lemma 2 : If $\phi:\Gamma\to 2^{\mathbb R}$ is such that
- $\forall x\in\Gamma$, $\phi(x)$ is a non-empty, closed interval
- $\forall x,y\in\Gamma$, $\phi(x+y)\subseteq \phi(x)+\phi(y)$
- $\forall x\in\Gamma,a\in[0,\infty[$, $\phi(ax)=a\phi(x)$
- $\forall x,y\in\Gamma$ with $x\leq y$, $\inf \phi(x)\leq\sup \phi(y)$
Then there exists a positive linear functional $f:X\to\mathbb R$ such that for all $x\in\Gamma$, $f(x)\in\phi(x)$.
The reason I would like to use this Lemma is that I would like to generalize some result relating to continuity that uses compactness of the dual $X'$ of $X$ to say that a net $\{ f_\alpha \}$ in it has a convergent subnet, and the $f$ it converges to has some desirable property. The goal is to try to reason about relaxation of this by defining $\phi(x)=[\liminf f_\alpha(x),\limsup f_\alpha(x)]$ (well if one of those is $\pm\infty$ we open the interval on that side). Now this $\phi$ satisfies the requirement of the Lemma (for istance $\liminf f_\alpha(x)+\liminf f_\alpha(y)\leq \liminf f_\alpha (x+y)$) and if the Lemma was true then the $f$ we obtain from it would satisfy everything I need. I now state the proof I attempted.
For two function $\mu,\nu$ satisfying the requirement of Lemma 2, we say that $\mu\preccurlyeq \nu$ if $\forall x\in\Gamma$, $\mu(x)\subseteq \nu(x)$. Let $A=\{ \mu : \mu\preccurlyeq \phi \}$, we will show that $A$ has a minimal element (using Zorn's Lemma) with the hope that this minimal element maps a singleton to each element of $\Gamma$, flattening such a function would give a function that satisfies assumptions of Lemma 1 and we would be able to conclude.
Let $\{ \mu_\alpha \}$ be a chain in $A$, such that $\alpha\leq \beta$ implies $\mu_\alpha\preccurlyeq \mu_\beta$. We show that $\nu$ defined as $\nu(x)=\bigcap_\alpha \mu_\alpha(x)$ is in $A$. It is quite clear that $\nu(x)$ is a non-empty closed interval because $\{ \mu_\alpha(x) \}$ forms a chain of non-empty closed interval ordered with $\subseteq$. Proving $\nu(ax)=a\nu(x)$ is trivial and so is the last requirement in Lemma 2. For the additivity like property : \begin{align*} \nu(x+y)&=\bigcap_\alpha \mu_\alpha(x+y)\\ &\subseteq \bigcap_\alpha (\mu_\alpha(x)+\mu_\alpha(y))\\ &=\nu(x)+\nu(y) \end{align*} Where the last line is due to the fact that if $a\in \bigcap_\alpha (\mu_\alpha(x)+\mu_\alpha(y))$, then for all $\alpha$, $(\mu_\alpha(x)-a)\cap \mu_{\alpha}(y)$ is non-empty. Now for any $\alpha,\beta$ we have that $(\mu_\alpha(x)-a)\cap \mu_{\beta}(y)$ is non-empty since one of $(\mu_\alpha(x)-a)\cap \mu_{\alpha}(y)$ or $(\mu_\beta(x)-a)\cap \mu_{\beta}(y)$ is contained in it (depending on if $\alpha\leq \beta$ or $\beta\leq \alpha$) and both are non-empty. Now if we fix $\beta$, we get that $\{ (\mu_\alpha(x)-a)\cap \mu_{\beta}(y) \}_\alpha$ is a chain of non-empty closed interval and so $A_\beta=\bigcap_\alpha (\mu_\alpha(x)-a)\cap \mu_{\beta}(y)$ is non-empty and closed, similarly $\{ A_\beta \}_\beta$ is a chain of non-empty closed interval and so $\bigcap_\beta A_\beta$ is non-empty, but \begin{align*} \bigcap_\beta A_\beta&=\bigcap_{\beta}\bigcap_\alpha (\mu_\alpha(x)-a)\cap \mu_{\beta}(y)\\ &=\left(\left(\bigcap_\alpha \mu_\alpha(x)\right)-a\right)\cap \left(\bigcap_{\beta}\mu_{\beta}(y)\right) \end{align*} and therefore $a\in \bigcap_\alpha \mu_\alpha(x)+\bigcap_{\beta}\mu_{\beta}(y)=\nu(x)+\nu(y)$.
We are now ready to apply Zorn's Lemma to get a minimal element $\psi$ in $A$. Here I cannot really finish the argument and I feel like there might be a need for another assumption on $\phi$. I wanted to argue by contradiction that if for some $x\in\Gamma$ we have $\psi(x)$ is not a singleton, then we would be able to pick $a\in\psi$ and build $\psi'$ such that $\psi'(x)=\{a\}$ and make sure that $\psi'$ satisfies the requirement and $\psi'\preccurlyeq \psi$ which would finish the proof. The construction for $\psi'$ I had in mind is \begin{align*} \psi'(y)=\bigcap_{b\in\mathbb R:y-bx\in\Gamma} [b\cdot a+\psi(y-b\cdot x)] \end{align*} I was able to show that this satisfies all requirement of Lemma 2 except the fact that $\psi'(x)$ is non-empty for all $x$. The problem I see here is that it could be that there is $y\in\Gamma$ such that $\psi'(x+y)\cap (a+\psi'(y))$ is empty, in this case $a$ is a bad choice and will yield $\psi'(x+y)=\emptyset$. I am wondering if without restriction on the requirement of Lemma 2, we can prove that there is a $a$ such that this problem doesn't occur, that would be enough to finish the proof, otherwise we somehow need to fix this in the Lemma (and hopefully keep it compatible with $\phi(x)=[\liminf f_\alpha(x),\limsup f_\alpha(x)]$ that I want to use it on).
Proof of Lemma 1 : We can trivially extend $L:\Gamma\to\mathbb R$ to some $L':\Gamma-\Gamma\to\mathbb R$ as $L'(x-y)=L(x)-L(y)$ for $x,y\in\Gamma$. Observe that $L'$ is linear and positive on $\Gamma-\Gamma\subseteq X$. In view of using the Hahn-Banach separation theorem, let us build a sublinear function on $X$ \begin{align*} p:X&\to\mathbb R\\ x&\to \max\left( L'(\overline x),-L'(\underline x) \right) \end{align*} One can easily check that $p$ is sublinear using the fact that $\overline{x+y}\leq \overline x+\overline y$ and $\overline{ax}=a\overline x$, together with positivity of $L'$. Also $L'\leq p$ on $\Gamma-\Gamma$, indeed for any $x\in\Gamma-\Gamma$, $x\leq\overline x$ implies $L'(x)\leq L'(\overline x)\leq p(x)$. By the Hahn-Banach separation theorem there exists a linear extension $f:X\to\mathbb R$ of $L'$ such that $f=L'$ on $\Gamma-\Gamma$ and $f(x)\leq p(x)$ on $X$. It remains to show that $f$ is positive, let $0\leq x$, since $p$ is a semi-norm, $-f(x-\overline x)\leq p(x-\overline x)$, but $x-\overline x\leq 0$ which means that $\underline{x-\overline x}\leq x-\overline x\leq \overline{x-\overline x}\leq 0$ and therefore $p(x-\overline x)=\max\left( L'\left(\overline{x-\overline x}\right),-L'\left(\underline{x-\overline x}\right) \right)=-L'\left(\underline{x-\overline x}\right)$. Therefore \begin{align*} f(x)&=-(-f(x-\overline x))+f(\overline x)\\ &\geq -p(x-\overline x)+L'(\overline x)\\ &=L'\left(\underline{x-\overline x}+\overline x\right) \end{align*} Now $\underline{-x}=-\overline x\leq x-\overline x$ since $0\leq x$ and so $-\overline {x}=\underline{\underline{-x}}\leq \underline{x-\overline x}$ which means that $0\leq \underline{x-\overline x}+\overline x$ and therefore $0\leq f(x)$, i.e. $f$ is positive.