In their proof of proposition $2$ of Carroll and Egorov (2019), they make an appeal to a theorem of the alternative that I don't quite understand. In particular, I don't see which version of the theorem they are appealing to and how it applies. I'd appreciate some assistance with this.
Let me try to explain the issue in a way that hopefully removes any need to look at the paper I've linked to.
Let $A=[0,\infty)^n$. For each $a\in A$ and $S \subset \{1,\ldots,n\}$, define $a\vert_S$ as the vector that coincides with $a$ along coordinates given in $S$, and is zero otherwise. That is, $a\vert_S = \hat a$, where $\hat a_i = a_i$ for $i \in S$, and $\hat a_i = 0$ otherwise.
We also fix a function $V:A\to\mathbb R$ that is weakly increasing in the product order. We normalise $V(0) = 0$. (I abuse notation here by also using $0$ to refer to the zero vector.) This implies that $V(a) \ge 0$ for all $a \in A$.
In the proof, they claim that, for each $a\in A$ with $V(a) > 0$, there exist non-negative numbers $r_1,\ldots,r_n$ such that $r_1 + \cdots + r_n = V(a)$, and that, for each subset $S \subset \{1 ,\ldots,n \}$, $$ \sum_{i\in S} r_i \le V \left( a \vert_S \right). $$
To show this, they argue by contradiction. The start of their argument reads
Suppose not. Then, applying a theorem of the alternative, we get the existence of nonnegative numbers $\lambda_S$, for each $S \subset \{1,\ldots,n \}$, such that $\sum_{S:i\in S}\lambda_S \ge 1$ for each $i$ and $\sum_S \lambda_S V \left( a \vert_S \right) < V(a)$.
Unfortunately, I don't quite see how the non-existence of appropriate $r_1,\ldots,r_n$ implies the claim given above. I am also not sure which version of the theorem they are relying on.
Can anyone help shed any light on this?
You can show existence of the numbers $(\lambda_{S})_{S}$ by appealing to the following variant of Farkas' lemma (see https://en.wikipedia.org/wiki/Farkas%27_lemma#): For $A\in\mathbb{R}^{m \times n}$ and $b\in\mathbb{R}^{m}$, either $Ax \leq b$ has a solution with $x \geq 0$, or $A^{T}y \geq 0$ has a solution with $b^{T}y < 0$ and $y \geq 0$.
Let $\iota = (1, \ldots 1) \in \mathbb{R}^{n}$ be the vector of ones. In the notation of the paper, the entries of the vector $\iota\vert_{S}$ are equal to one only at entries corresponding to elements of $S$.
Let's enumerate the proper subsets of $\lbrace 1, \ldots, n\rbrace$ as $S_{1}, \ldots, S_{2^{n}-1}$. We may assume that $V(a) > 0$ holds (as otherwise one could simply choose $r_{i} = 0$ for all $i$). Now let the matrix $A$ be $$ A= \begin{pmatrix} - \iota^{T} / V(a) \\ \left(\iota \vert_{S_{1}}\right)^{T} \\ \vdots \\ \left(\iota \vert _{S_{2^{n}-1}}\right)^{T} \end{pmatrix}, $$ and let $b$ be the vector $$ b = \begin{pmatrix} - 1 \\ V\left(a\vert_{S_{1}}\right) \\ \vdots \\ V\left(a\vert_{S_{2^{n}-1}}\right) \end{pmatrix}. $$
If the numbers $r_{1}, \ldots r_{n}$ do not exist, then, in particular, there is no vector $r$ such that $$ Ar \leq b $$ Farkas' lemma therefore implies existence of non-negative numbers $\left(\lambda_{0}, \lambda_{1}, \ldots, \lambda_{2^{n}-1}\right)$ such that $A^{T} \lambda \geq 0$ and $b^{T} \lambda < 0$. The first inequality asserts that for all $i \in\lbrace 1, \ldots, n\rbrace$, $$ - \frac{\lambda_{0}}{V(a)} + \sum_{j \colon i\in S_{j}, j\neq 0} \lambda_{j} \geq 0 $$ holds. The second inequality is nothing but $$ - \lambda_{0} + \sum_{j \neq 0}V\left(a\vert_{S_{j}}\right) \lambda_{j} < 0. $$ Notice that the second inequality implies $\lambda_{0} > 0$. Hence, if we normalize by setting $\lambda_{j}^{\prime} = \lambda_{j} \lambda_{0} / V(a)$ for all $j \neq 0$, the numbers $\left(\lambda_{j}^{\prime} \right)_{j}$ give us what we want.