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For fixed $x$, put $z=r-r_0$. The polyhedral uncertainty set is equivalent to $-1\le Pz\le 1$, so the inner problem is the linear programThe variable $z$ is unrestricted in sign. The point $z=0$ is strictly feasible. If the optimum is finite, it is attained, so the infimum is a minimum. The use of an infimum also covers possible unbounded cases permitted by the printed assumptions; strict feasibility alone does not guarantee a finite objective.

Worst-case linear revenue over a polyhedral uncertainty set is optimized by the linear programHere ${\Delta }_n$ is the probability simplex. Finite decisions are exactly its intersection with $\mathrm{range}{P}^T$. If this intersection is empty, all decisions have worst-case value $-\infty$ and the displayed linear program is infeasible; its supremum over the empty feasible set is also $-\infty$. Full column rank of $P$ suffices to make all simplex decisions finite.

If $x\notin \mathrm{range}{P}^T=(\ker P{)}^{\perp }$, the support function is infinite along a kernel line. Otherwise linear programming duality gives the displayed minimum, attained because the corresponding linear programs are feasible with finite values. Splitting $q=\alpha -\beta$ into nonnegative parts yields the equivalent minimum of $1^T(\alpha +\beta )$. Thus worst-case revenue over a polyhedral uncertainty set is $r_0^{T}x-{\sigma }_C(x)$, with value $-\infty$ off the row space.