Active constraint set 2026-10-31
For inequalities , the active constraints at a feasible point are those holding with equality. Inactive constraints remain strictly satisfied in a sufficiently small neighborhood and do not restrict first-order feasible velocities there. The active gradients determine the linearized tangent cone and appear in the linear independence constraint qualification. At a KKT point, inactive constraints have zero Lagrange multipliers by complementary slackness.
Contingent tangent cone 2026-10-31
This cone records first-order limiting velocities of feasible points approaching . The defining limits are required to exist, and the point sequences must be feasible. A local minimizer of a differentiable function satisfies for every in the cone: expand to first order and divide by . A tangent direction need not itself generate an exactly feasible straight segment. Linearizing active constraints always gives a containing linearized tangent cone; equality requires a suitable constraint qualification.
The active constraint set is , and the linearized tangent cone isDifferentiating each active feasibility inequality along a tangent sequence gives . A sufficient condition for equality is the linear independence constraint qualification: the active constraint gradients are linearly independent. The weaker Mangasarian–Fromovitz constraint qualification also suffices in this inequality-only setting.
Under the stated independence condition, the necessary first-order conditions become the Karush-Kuhn-Tucker conditions: there are multipliers withUse the allowed span assertion to write . Independence gives vectors with for the active indices. Each lies in , so part (a) gives . Assign zero multipliers to inactive constraints. This proves all displayed conditions, including complementarity. Conversely, their stationarity and sign conditions imply for every . Thus they reformulate the first-order inequality; they do not alone prove a minimum for a nonconvex problem.