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Associate nonnegative Lagrange multipliers $\alpha$ with $-Pz-1\le 0$ and $\beta$ with $Pz-1\le 0$. The Lagrangian dual problem is obtained fromIts infimum over unrestricted $z$ is finite exactly when $P^T(\alpha -\beta )=x$. Thus the dual isThe precise finiteness condition is $x\in \mathrm{range}{P}^T$. If $x=P^{T}q$, then $x^{T}z=q^{T}Pz\ge -\Vert q{\Vert }_1$ on the feasible set. The primal is feasible and bounded below, and splitting $q$ into positive and negative parts makes the dual feasible. Linear programming duality therefore gives attained equal finite optima. Equivalently, the strictly feasible point $z=0$ satisfies the Slater condition, with the required finiteness hypothesis. This is strong duality in the usual finite sense.

If $x\notin \mathrm{range}{P}^T=(\ker P{)}^{\perp }$, there is $d\in \ker P$ with $x^{T}d<0$. Every $z=td$, $t\ge 0$, is feasible and the objective tends to $-\infty$. The dual is then infeasible. Equality of extended values still holds if its supremum over the empty feasible set is defined as $-\infty$, but there are no finite optimizers. The paper does not assume full column rank of $P$, so this qualification is necessary. The same calculation gives the support function of an inverse image of an infinity-norm ball, ${\inf }_r{r}^{T}x=r_0^{T}x-{\min }_{P^{T}q=x}\Vert q{\Vert }_1$ for finite decisions.