Exercise 7, § 3

Let T be a commutative monoid (A, I, § 2.1) carrying a preorder relation $x \leq y$ such that if $x \leq y$ then $x + z \leq y + z$ for all $z \in T$. A mapping f of T in $\mathbf{R} \cup \{ -\infty \}$ is called additive (resp. subadditive, resp. superadditive) if we have

for any $x, y$ in T. a) If g is subadditive and increasing in T, then the function $h(x) = \inf_{n > 0} g(nx)/n$ is subadditive and increasing; we have $h \leq g$ and $h(0) = 0$ if $g(0) \geq 0$. b) Under the same hypotheses suppose that there exist two elements $x_1, x_2$ of T and two real numbers $\xi_1, \xi_2$ such that $\xi_1 < g(x_1), \xi_2 < g(x_2)$ and $g(x_1 + x_2) < \xi_1 + \xi_2$. Let $y_1, y_2, z_1, z_2$ be four elements of T, let $n_1, n_2$ be two integers $\geq 0$ and let $\alpha_1, \alpha_2$ be two real numbers such that Show that then $g(n_2 y_1 + n_1 y_2) < n_2 \alpha_1 + n_1 \alpha_2$. c) Let $\omega$ be a superadditive function on T such that $\omega(0) = 0$, and let $\Omega$ be an increasing subadditive function on T such that $\omega(x) \leq \Omega(x)$ in T. Show that there exists an increasing additive function f on T such that $\omega(x) \leq f(x) \leq \Omega(x)$ in T. (Remark that the set of increasing subadditive functions g on T such that $\omega(x) \leq g(x) \leq \Omega(x)$ in T is non-empty and inductive for the relation $\geq$, and take a minimal element of this set for f; show using a) that $f(0) = 0$. To show that there cannot exist pairs of elements of T, $(x_1, x_2)$ such that $f(x_1 + x_2) < f(x_1) + f(x_2)$ remark that if $\xi_j \in \mathbf{R}$, and $h_j(x) = \inf(n \xi_j + f(y))$ where n varies in the set of integers $\geq 0$ and y in the set of elements of T such that $x \leq nx_j + y$, then $h_j$ is increasing and subadditive in T ($j = 1, 2$), $h_j(x_j) \leq \xi_j$ and $h_j(x) \leq f(x)$ for all $x \in T$. Then use the definition of f and part b) to obtain a contradiction.)