Function classes on posets, which are closed under certain operations

Let $(S,\leq)$ be a poset. Let $\Omega = \{f : S \to \mathbb R : f \text{ is monotone decreasing}\}$ be the set of all decreasing functions i.e. those such that $f(x) \leq f(y)$ for all $x \geq y$. Then, $\Omega$ has a few desirable properties :

• $0 \in \Omega$ (The identically zero function).

• For $f,g \in \Omega$, we have $f+g \in \Omega$.

• For $f,g\in \Omega$, we have $\max\{f,g\} \in \Omega$ and $\min\{f,g\} \in \Omega$.

• $\Omega$ is closed under pointwise convergence of functions.

These properties are desirable to me, because I am performing an iterative procedure like so : let $\mathbb R^{ S} = \{f : S \to \mathbb R\}$. Given a linear operator $P : \mathbb R^ S \to \mathbb R^{ S}$, I write up sufficient conditions on $P$ so that $P:\Omega \to \Omega$ ($P$ involves the maximum, minimum and sum operations). Then, by beginning with some $f_0 \in \Omega$, and defining $f_n = Pf_{n-1}$, I ensure that (using some Banach FPT procedure) the $f_n$ converge to $f \in \Omega$ such that $Pf = f$. This $f$ is a fixed point with structure of $P$.

I wish now, to find similar sets like $\Omega$, so that with conditions for $P$ to be preserving that class (and hoping $f_0$ belongs in that class), I can find and prove the existence of various structured fixed points of $P$.

Question : Can somebody come up with more kinds of subsets of $\mathbb R^{S}$ such that each of the properties listed above (for $\Omega$) is also satisfied by that class?

Here are some non-examples : convex and concave functions don't fall in here because of the $\min$ , $\max$ non-closure respectively. These are important classes ruled out, but I am hoping there are still enough examples around.

Note : If anybody wants a concrete poset to work on, please consider $\mathbb N^3$ with the lattice partial order. Answers restricted to just this poset are also acceptable.

Further context : The function $f$ satisfying $Pf=f$ is a "two policy comparison" function. So, I have an optimal control problem where I must compare two policies, and $f$ describes the difference between these policies pointwise. Providing structure on $f$ (more precisely on the sign of $f$) gives a structure to when one policy is preferred to the other : an important example of this is monotone policies in queue theory (where $f$ is , in fact monotone decreasing). This is how this problem links to optimal control.

Solution 1:

There are ridiculously many examples; far too many to classify. Here's a rich family I encountered on my attempt at classification.

First, let $\{G_s\}_{s\in S}$ be any sequence of closed submonoids of $(\mathbb{R},+)$. This includes $0$ and any discrete subgroup, so we already have at-least-continuum-many possibilities. Second, pick $c\in S^2\to\mathscr{P}(\{\leq,\geq\})$ where $\mathscr{P}$ is the power-set operator. Then it is easy to verify that $$V(G,c)=\{f:{{R\in c(s,t)}\implies f(s)\mathrel{R}f(t)}\}\subseteq\prod_{s\in S}{G_s}$$ satisfies your conditions.

There are still more examples; the key problem is that we can "lose track" of which increments correspond to which functions. For example, suppose $S=[2]$ and $$\mathcal{X}=\{f:{f=0}\vee f\text{ strictly increasing}\}$$ If $\mathcal{X}\subseteq V(G,c)$, then $G=(\mathbb{R},\mathbb{R})$ and $c(1,2)={\leq}$. But $(1,1)\in V(G,c)\setminus\mathcal{X}$. One might hope to rectify this, by make $c$ not just an order relationship, but a full additive monoid of possible differences. I still don't think that's enough to classify, but I can't prove it.

Moreover, you place no conditions on $S$; you probably want to restrict $S$ to be at-most-countable. Otherwise, pointwise convergence begins to break down, since sequences are only countably long: taking the pointwise closure of the vector space generated by the standard basis $\{\delta_s\}_{s\in S}$ does not give $\mathbb{R}^S$, but rather $\{f:\mathrm{supp}{f}\text{ countable}\}$.