Let an upper density (on $\mathbf N$) be a (set) function $f: \mathcal P(\mathbf N) \to \mathbf R$ such that, for all $X, Y \subseteq \bf N$ and $h,k \in \mathbf N^+$, the following hold:



> 
> (F1) $f(\mathbf N) = 1$;
> 
> 
> (F2) $f(X) \le f(Y)$ whenever $X \subseteq Y$;
> 
> 
> (F3) $f(X \cup Y) \le f(X) + f(Y)$;
> 
> 
> (F4) $f(k \cdot X + h) = \frac{1}{k} f(X)$, where $k \cdot X + h :=
> \{kx+h: x \in X\}$.
> 
> 
> 


It was noted in another thread ([here](https://mathoverflow.net/questions/214878)) that an upper density $f$ doesn't need to be weakly additive, meaning that $f(X \cup Y) = f(X) + f(Y)$ for all disjoint $X, Y \subseteq \bf N$ such that $Y$ is an (infinite) arithmetic progression (namely, a set of the form $k \cdot \mathbf N + h$ for some $h \in \mathbf N$ and $k \in \mathbf N^+$).


There are, however, a bunch of upper densities that *are* weakly additive, which is notably true of the upper logarithmic, upper asymptotic, and upper Banach densities (along with at least uncountably many others).


In fact, upper densities form a convex subset, $\mathscr U$, of the real vector space, $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$, of bounded functions $\mathcal P(\mathbf N) \to \bf R$, so it makes sense to ask how the [extreme points](https://en.wikipedia.org/wiki/Extreme_point) of $\mathscr{U}$, relative to the linear structure of $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$, look like. Hence my question:



> 
> **Q.** Is it true that every extreme point of $\mathscr{U}$ is a weakly additive upper density? (The answer might depend on the axiom of choice, so let us assume to work in ZFC.)
> 
> 
> 


Here are some positive results: If $\preceq$ denotes the (partial) order on the set, $\hom(\mathcal P(\mathbf N), \mathbf R)$, of all functions $\mathcal P(\mathbf N) \to \bf R$ defined by $f \preceq g$ iff $f(X) \le g(X)$ for all $X \subseteq \bf N$, then it is seen that $\mathscr{U}$ is a complete subsemilattice of the join-semilattice $(\hom(\mathcal P(\mathbf N), \mathbf R), \preceq)$, meaning that, whenever $\mathscr{D}$ is a nonempty subset of $\mathscr{U}$, the set $$\{u \in \mathscr U: f \preceq u\text{ for all }f \in \mathscr{D}\}$$ has a least element, relative to the order $\preceq$, that still belongs to $\mathscr{U}$. In particular, $\mathscr{U}$ has a maximum (again, relative to the order $\preceq$), which is given by the upper Buck density (on $\bf N$), that is the function
$$
\mathfrak{b}^\ast: \mathcal P(\mathbf N) \to \mathbf R: X \mapsto \inf\_{S \in \mathscr{A}: X \subseteq S} \mathsf{d}^\ast(S),
$$
where $\mathscr{A}$ is the set of all subsets of $\mathbf N$ that can be written as a finite union of arithmetic progressions, and $\mathsf d^\ast$ is the upper asymptotic density (on $\bf N$).


It is found that $\mathfrak b^\ast$ is a weakly additive upper density, and on the other hand, it can be proved, under the axiom of choice, that $\mathscr{U}$ has at least uncountably many minimal elements, relative to the order $\preceq$ (I don't know if the existence of even one minimal point can be proved in ZF).


Now, it is not difficult to prove that minimal and maximal elements of $\mathscr{U}$ are extreme points of $\mathscr{U}$; unfortunately, it is not clear to me if also the converse is true (if yes, the answer to question Q would be in the affirmative), though I don't *believe* so.



Are the extremal points of a certain set of functions $\mathcal P(\mathbf N) \to \bf R$ weakly additive?
I think that the answer is **No**.


The space $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$ with pointwise convergence is a [locally convex space](https://en.wikipedia.org/wiki/Locally_convex_topological_vector_space). (It can be considered a subspace of $\mathbf R^{\mathcal P(\mathbf N)}$.)


The set $\mathscr U$ is compact in this space. (It can be identified with a closed subset of $[0,1]^{\mathcal P(\mathbf N)}$.)


So by [Krein-Milman theorem](https://en.wikipedia.org/wiki/Krein%E2%80%93Milman_theorem) we get that $\mathscr U$ is closed convex hull of the set of all extreme points of $\mathscr U$.


On the other hand, the set of all weakly additive upper densities is closed and convex subset of $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$. (It suffices to check that limit of a convergent net of weakly additive upper densities is again a weakly additive upper density and that a convex combination of weakly upper densities is a weakly upper density.) So if all extreme points of $\mathscr U$ were weakly additive, the same would be true for the whole set $\mathscr U$.


