Lemma 6

We maintain the hypotheses and notations of Lemma 5, and assume the $f_\lambda$ to be positive. The function $\overline{f}$ is positively homogeneous, concave and upper semi-continuous in $C$. It is affine if $C$ is lattice-ordered.

It is clear that $\overline{f}$ is positively homogeneous. Let $x, y$ belong to $C$. If $x_1, \ldots, x_m, y_1, \ldots, y_n$ in $C$ are such that $x_1 + \cdots + x_m = x,\ y_1 + \cdots + y_n = y$, then $x_1 + \cdots + x_m + y_1 + \cdots + y_n = x + y$, therefore

it follows that $\overline{f}(x) + \overline{f}(y) \leq \overline{f}(x + y)$, therefore $\overline{f}$ is concave. Let $L$ (resp. $L_\lambda$) be the set of $(t, x) \in \mathbf{R} \times E$ such that $x \in C$ and $0 \leq t \leq \overline{f}(x)$ (resp. $0 \leq t \leq f_\lambda(x)$). Each of the $L_\lambda$ is closed in the weakly complete proper convex cone $\mathbf{R}_+ \times C$, therefore the sum $\sum_{\lambda \in \Lambda} L_\lambda$ is closed (TVS, II, §6, No. 8, Cor. 2 of Prop. 11). By Lemma 5, this sum is equal to $L$. Therefore $L$ is closed, which proves that $\overline{f}$ is upper semi-continuous. Finally, assume $C$ is lattice-ordered, and let us prove that $\overline{f}$ is convex. Let $x, y$ belong to $C$ and let $\varepsilon > 0$. There exist $z_1, z_2, \ldots, z_n$ in $C$ such that $f(z_1) + \cdots + f(z_n) \geq \overline{f}(x + y) - \varepsilon$ and $z_1 + \cdots + z_n = x + y$. The vector space $C - C$ is lattice-ordered for the order induced by that of $E$ (A, VI, §1, No. 9, Prop. 8). By the decomposition theorem (loc. cit., No. 10, Th. 1), there exist $x_1, \ldots, x_n, y_1, \ldots, y_n$ in $C$ such that

Then, since $f$ is positively homogeneous and convex,

Since $\varepsilon$ is an arbitrary number $> 0$, we have proved that $\overline{f}$ is indeed convex.