Separate multiplier necessity from convex sufficiency

Try

Minimize (x−4)² subject to x≤2. The best feasible point is x=2, where the objective derivative is −4. Write the inequality as g(x)=x−2≤0 and the Lagrangian as L=f+λg. Stationarity requires −4+λ=0, so λ=4. Its nonnegative sign reflects a minimization problem with upper-form inequalities. The active boundary blocks the direction in which the objective would otherwise decrease.

Predict

Predict the multiplier if the unconstrained minimizer becomes feasible, for example minimizing (x−1)² under the same constraint. At x=1 the inequality is slack and λ=0. Complementary slackness λg=0 captures both cases. Predict also whether convexity alone guarantees such multipliers exist. It does not: a badly degenerate representation can prevent stationarity, even when the feasible point is a global minimum of a convex problem.

Experiment

Compare the two quadratic objectives under x−2≤0. Check all four conditions, not just the derivative equation.

Objective	Candidate x	Constraint g	Multiplier λ	Stationarity
(x−4)²	2	0	4	−4+4=0
(x−1)²	1	−1	0	0+0=0

In both rows primal feasibility, dual nonnegativity and λg=0 hold. Now try x=2,λ=−4 in the second row: even if an altered arithmetic expression appeared attractive, the wrong multiplier sign would invalidate the convex lower-bound argument. Keep the maximization LP convention from earlier lessons separate from this minimization Lagrangian convention.

Understand

For differentiable minimization with gᵢ(x)≤0 and hⱼ(x)=0, the Karush–Kuhn–Tucker conditions are primal feasibility, λᵢ≥0, stationarity ∇f+Σλᵢ∇gᵢ+Σνⱼ∇hⱼ=0, and complementary slackness λᵢgᵢ=0. Equality multipliers ν are free. A sufficient qualification for local necessity is LICQ: the gradients of equalities and active inequalities are linearly independent at the point, with the usual local smoothness assumptions. The multiplier theorem uses the geometry of linearized feasible directions and a separation argument; independence prevents a degenerate abnormal relation from replacing the objective gradient. This is the proof mechanism, not a claim that checking stationarity proves LICQ.

For convex f and gᵢ with affine equalities, a feasible KKT point is globally optimal without needing a qualification for that sufficiency direction. The Lagrangian is convex because λ≥0. Stationarity makes the candidate its global minimizer. For any feasible x, L(x,λ,ν)≤f(x), while complementary slackness makes L at the candidate equal its objective. The resulting lower bound certifies global optimality. A qualification is needed to guarantee existence of multipliers at an optimum, not to validate an already verified convex certificate.

Apply

Minimize x subject to x²≤0. The only feasible point is zero and the problem is convex. Yet stationarity would be 1+2λx=1=0 at that point. LICQ fails because the active gradient is zero, and strict feasibility also fails. This example prevents the false inference “convex optimum, therefore KKT multipliers automatically exist.” It will later illustrate dual nonattainment even when the primal and dual optimal values agree.

For a nonconvex problem, KKT can hold at a local maximum or saddle. Add second-order analysis or a global certificate rather than promoting necessary equations into sufficient conditions. Numerical KKT residuals also need a scale: a tiny stationarity residual alongside large primal infeasibility does not establish an approximately feasible solution.

Continue

Write the four KKT checks and the logical role of any qualification separately. Continue to Second-order constrained conditions, where curvature is tested only in relevant feasible directions. Constraint curvature enters the Lagrangian Hessian, explaining why the objective Hessian alone can miss a strict constrained minimum.