Theorem 8.11 — with $K_2$ polyhedral and finite value, the deterministic equivalent convex program (8.2) is stable

Consider a stochastic program with fixed recourse whose random element $\xi$, with law a probability measure $\mu$, satisfies the weak covariance condition (Definition 2.2), and whose recourse matrix $W$ has full row rank (standing assumption, p. 312). Let $Z(x)=\bar cx+\mathcal Q(x)$ be the objective (3.1) and $K_2$ the induced constraint set (Corollary 4.5). Consider the deterministic equivalent convex program (8.2),

Suppose that

1. $K_2$ is a convex polyhedron, and
2. the program is finite: $v=\inf\{Z(x): x\in K_1\cap K_2\}$ is a real number.

Then (8.2) is stable: there is a multiplier $\pi\in\mathbb R^m$ for the fixed constraints $Ax=b$ with

Equivalently, the dual program (8.3), $\sup\{u \mid u\le\mathcal Q(x)+(\bar c-\pi A)x+\pi b\ \ \forall x\in K_2\cap\{x\ge0\}\}$, is solvable and there is no duality gap.

Stability means that the optimal value reacts at a bounded rate to perturbations $b\mapsto b-u$ of the first-stage constraints. The paper's Example 8.5 shows that the weak covariance condition cannot be dropped.

Formalization Note The paper states the hypothesis as "$K$ is polyhedral" with $K=K_1\cap K_2$. Read literally, that is false: with $K_2$ the closed unit disk in $\mathbb R^2$ (produced by $T(\xi)$ uniform on the unit circle, $p\equiv1$, $W=(1)$, $q\equiv0$), $c\equiv(-1,0)$, and $K_1=\{x_2=1,\ x\ge0\}$, the set $K=\{(0,1)\}$ is polyhedral and the value $0$ is finite, but $-x_1+\pi(1-x_2)<0$ at $x=(\sqrt{2h-h^2},1-h)$ for small $h>0$, so no multiplier exists. The paper's own sentence before Lemma 8.9 ties "polyhedral region" to $K_2$ (Theorems 4.7, 4.10 concern $K_2$), and the proof applies Lemma 8.9 with the objective's domain $K_2$; the statement here uses $K_2$. In (8.3) the page writes $c$ where (3.1) and (8.2) require $\bar c$. The middle constraint of (8.2), written with the polar matrix $W^*$, is the constraint $x\in K_2$ by Corollary 4.5.