The Lagrange Dual

Price the constraints instead of enforcing them, and a concave lower bound appears under any problem at all.

The idea

Lagrange duality attaches a price to each constraint of an optimization problem and, from the prices, produces lower bounds on its optimal value.

Attach a multiplier $\lambda_{i} \ge 0$ to each inequality constraint $f_{i}(x) \le 0$ and $\nu_{j}$ to each equality $h_{j}(x) = 0$. Instead of enforcing a constraint, charge $\lambda_{i} f_{i}(x)$ or $\nu_{j} h_{j}(x)$ for violating it. The objective plus the charges is the Lagrangian

$L(x, \lambda, \nu) = f_{0}(x) + \sum_{i=1}^{m} \lambda_{i} f_{i}(x) + \sum_{j=1}^{p} \nu_{j} h_{j}(x).$

At fixed prices, the smallest total any point can achieve is the Lagrange dual function

$g(\lambda, \nu) = \inf_{x} L(x, \lambda, \nu),$

the infimum running over the whole domain, feasible points and infeasible ones alike.

Two facts hold with no assumptions on the primal problem — not convexity, not even continuity.

Theorem (Weak duality).

Let $p^{\star}$ be the optimal value of the primal problem. Then the dual function $g$ is concave, and $g(\lambda, \nu) \le p^{\star} \quad \text{for every } \lambda \succeq 0 \text{ and every } \nu.$

The dual problem asks for the best of those bounds: maximize $g(\lambda,\nu)$ subject to $\lambda \succeq 0$. Its optimal value $d^{\star}$ satisfies $d^{\star} \le p^{\star}$, and the difference $p^{\star} - d^{\star}$ is the duality gap. In LP Duality the dual of a linear program was another linear program; the Lagrangian builds a dual for any problem at all.

Ways to work on it

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