4
$\begingroup$

Let $\mu^\star$ be a real-valued function defined on the power set of the positive integers $\mathbf{N}^+$ such that for all $X,Y\subseteq \mathbf{N}^+$ the following axioms hold:

(F1) $\mu^\star(\mathbf{N}^+)=1$;

(F2) $\mu^\star(X) \le \mu^\star(Y)$ if $X\subseteq Y$;

(F3) $\mu^\star(X\cup Y) \le \mu^\star(X)+\mu^\star(Y)$;

(F4) $\mu^\star(\{kx:x \in X\})=\mu^\star(X)/k$ for all $k \in \mathbf{N}^+$;

(F5) $\mu^\star(\{x+h: x \in X\})=\mu^\star(X)$ for all $h \in \mathbf{N}^+$.

A function of this type is said to be an (arithmetic) upper density. The set of these functions include the upper asymptotic, Banach, logarithmic, analytic, Polya, Buck densities and many others. Related questions on these type of functions can be found here and here.

At this point we can defined its associated lower density $\mu_\star$ for all $X\subseteq \mathbf{N}^+$ by $$\mu_\star(X)=1-\mu^\star(X^c).$$ Now, all examples I have in mind are superadditive functions, namely $$ \mu_\star(X\cup Y) \ge \mu_\star(X)+\mu_\star(Y) $$ whenever $X$ and $Y$ are disjoint subsets of $\mathbf{N}^+$. Does this property hold in general?

After an easy manipulation, the question turns out to be equivalent to the following:

Question. Let $\mu^\star$ be an upper density on $\mathbf{N}^+$, that is, a function satisfying axioms (F1)-(F5). Is it true that if $X,Y$ are subsets of $\mathbf{N}^+$ such that $X\cup Y=\mathbf{N}^+$ then $$ 1+\mu^\star(X\cap Y) \le \mu^\star(X)+\mu^\star(Y)? $$


In turn, this can be viewed as a strenghtening of (F3) above; moreover, it would imply that the induced density $\mu$, which can be seen as the restriction of $\mu^\star$ on $\{X\subseteq \mathbf{N}^+:\mu^\star(X)=\mu_\star(X)\}$, is additive.

$\endgroup$
2
$\begingroup$

I will reuse the same trick as in the answer to Paolo's other question to show that the answer to this is still in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h := ((1-\alpha) f^q + \alpha g^q)^{\frac{1}{q}}$$ is an upper density too (most notably, condition (F3) follows from Minkowski's inequality, which is where we need $q \ge 1$). So in particular, assume from now on that $f$ is the upper asymptotic (or natural) density (on $\mathbf N^+$), namely the function $$ \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \limsup_{n \to \infty} \frac{|X \cap [1,n]|}{n}, $$ and $g$ the upper Banach (or uniform) density, viz. the function $$ \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \lim_{n \to \infty} \max_{h \ge 0} \frac{|X \cap [1+h,n+h]|}{n} $$ (the latter limit exists as a consequence of Fekete's lemma); both of these functions are upper densities, as was mentioned in the OP.

Next, fix $a \in {]0,1]}$, and set $$ V_a := \bigcup_{n \ge 1} [\![a(2n-1)!+(1-a) (2n)!, (2n)!]\!] $$ and $$ X := V_a \cup (V_a^c \cap (2\cdot \mathbf N)) \quad\text{and}\quad Y := V_a \cup (V_a^c \cap (2 \cdot \mathbf N+1)), $$ where $V_a^c := \mathbf N^+ \setminus V_a$. It is clear that $X \cup Y = \mathbf N^+$ and $X \cap Y = V_a$. Moreover, we get from here that $$ f(X) \le f(V_a) + f(V_a^c \cap (2 \cdot \mathbf N)) \le a +\frac{1}{2}, $$ and similarly $f(Y) \le a + \frac{1}{2}$ and $f(V_a) = a$. On the other hand, $V_a$ contains arbitrarily large intervals of consecutive integers, hence $g(X) = g(Y) = g(V_a) = 1$. It follows that $$ 1+h(X \cap Y) = 1 + (\alpha +a^q(1-\alpha))^{\frac{1}{q}} $$ and $$ h(X) + h(Y) \le 2 \cdot (\alpha +(0.5+a)^q(1-\alpha))^{\frac{1}{q}}. $$ With this in hand, suppose to a contradiction that $1+h(A \cup B) \le h(A) + h(B)$ for all $A, B \subseteq \mathbf N^+$ such that $A \cup B = \mathbf N^+$, regardless of the actual values of the parameters $a$, $\alpha$ and $q$. Then, we have from the above that $$ 1 + \alpha^{\frac{1}{q}} \le 2 \cdot (\alpha +\left(0.5+a\right)^q(1-\alpha))^{\frac{1}{q}}, $$ which ultimately implies, in the limit as $a \to 0^+$, that $$ 1 + \alpha^{\frac{1}{q}} \le (2^q\alpha + 1-\alpha)^{\frac{1}{q}} $$ for all $\alpha \in [0,1]$ and $q \in [1,\infty[$. This, however, is quite false (e.g., let $\alpha = \frac{1}{2}$ and $q = 2$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.