I'm perusing this paper: ftp://ftp.cs.wisc.edu/pub/dmi/tech-reports/13-01.pdf
and I'm stuck in how to deduce this equation:
$v'(x + y) + w'(−x + y) = 0$
from this optimization problem: $$\min_{\substack{x,y}} h'y$$
s.t. $$Ax − y = b$$ $$x + y \geq 0$$ $$−x + y \geq 0$$
and its dual:
$$\max_{\substack{u, v, w}} b'u$$
s.t. $$A'u + v − w = 0$$ $$−u + v + w = h$$ $$v\geq 0$$ $$w\geq 0$$
I think the author uses the Strong Duality, as long as the set $\mathbb{R^{nxm}}$ is convex and the problem has a feasible solution.
I asked this question to my teacher and it became clear. Paper's author uses the KKT conditions. In the primal minimization problem we have:
$Ax − y - b = 0$ it is lineal affine.
$-x - y \leq 0$ convex
$x - y \leq 0$ convex
The domain is all the space $\mathbb{R}^n$, then this problem meet the Slater's conditions, so I can apply KKT using $v\geq 0$ and $w\geq 0$ such as: $$v'(-1)(x+y) = 0$$ $$w'(-1)(-x+y) = 0$$ then $$v'(x+y) + w'(-x+y) = 0$$