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Lesson 19 of 27

Inspect curvature in feasible critical directions

Use the Lagrangian Hessian rather than the objective Hessian alone.

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Minimize the linear objective f(x,y)=x on the unit circle x²+y²=1. Its global minimum is at (−1,0), although the objective Hessian is identically zero. Use L=x+λ(x²+y²−1). Stationarity at the candidate gives λ=1/2, so the Lagrangian Hessian is the identity matrix. On the vertical tangent direction it is positive. The constraint’s curvature, weighted by its multiplier, supplies the second-order information missing from the objective alone.

Predict

Predict what happens at the opposite point (1,0). Its multiplier is −1/2 and the Lagrangian Hessian is minus the identity, giving negative tangent curvature and a maximum. Predict whether curvature should be checked in every ambient direction. It should not: directions leaving the feasible surface at first order are irrelevant to the local equality-constrained test. The tangent restriction is essential, not a weakened approximation chosen to make an answer pass.

Experiment

Parametrize the circle near its left point by x(t)=−cos t, y(t)=sin t. Its initial velocity is (0,1), and the objective is −cos t=−1+t²/2+terms of higher order.

CandidateMultiplierTangent vectorLagrangian quadratic value
(−1,0)1/2(0,1)1
(1,0)−1/2(0,1)−1

The coefficient in the Taylor expansion is half the displayed quadratic value, matching the standard second-order formula. Check this factor rather than confusing a second derivative with the quadratic coefficient itself. Both rows satisfy first-order stationarity, so the sign distinguishes their local behavior.

Understand

For smooth regular equality constraints h=0, take a feasible curve x(t) with x(0)=x⋆ and velocity d. Differentiating the constraints gives Dh d=0; differentiating again gives Dh x″ plus the vector of constraint Hessian quadratic forms equal to zero. The second derivative of f along the curve is dᵀ∇²f d+∇f·x″. Substitute stationarity ∇f=−Dhᵀλ and the second constraint derivative. The result is dᵀ∇²L d. Thus a local minimum requires nonnegative Lagrangian curvature on every tangent direction under the regularity assumptions ensuring such curves exist.

Conversely, if Dh has full row rank, stationarity holds, and ∇²L is positive definite on every nonzero tangent direction, a local constraint chart and Taylor expansion prove a strict local minimum. Positive definiteness in the whole ambient space is unnecessary. The reduced Hessian Zᵀ∇²L Z, where columns of Z form a tangent basis, makes the test computational. Its positive definiteness is independent of the particular full-rank basis used for that same tangent space.

Apply

For inequalities, critical directions satisfy linearized equalities, nonpositive directional derivatives of active inequalities, and zero first-order objective change. Under appropriate qualifications, necessary second-order tests focus on this critical cone. Directions with a strictly positive objective derivative already increase the objective at first order. At a KKT point, active constraints with positive multipliers impose equality along a critical direction; active constraints with zero multipliers can retain one-sided freedom. Treating every active inequality as an equality can therefore omit relevant directions.

The full inequality second-order theorem needs its qualification and multiplier hypotheses; this lesson uses the equality case as the completely derived core and states the critical-cone extension with its limits. A semidefinite reduced Hessian remains inconclusive: higher-order terms may determine the outcome, just as in unconstrained x⁴ versus x³ examples. Numerical near-zero eigenvalues also require a tolerance and conditioning assessment.

Continue

Keep the feasible tangent geometry, multiplier convention and Lagrangian Hessian together in a proof. Continue to Convex duality and Slater, which constructs lower bounds directly from the Lagrangian. In a convex problem, global lower-bound reasoning can certify optimality without a local second-order classification, but existence of a dual certificate still needs attention to qualification and attainment.

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Practice and reasoning

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For min x²+3y² subject to x+y=0, evaluate dᵀ∇²f d for d=(1,−1).

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

For min x subject to x²+y²=1 at (−1,0), using L=x+λ(x²+y²−1), what is λ?

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

State the second-order necessary condition under regular smooth equality constraints.

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Explain why the objective Hessian alone misses the strict minimum of x on the unit circle at (−1,0).

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For g≤0, define critical directions at a KKT point and explain their role.

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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

State a standard equality-constrained sufficient condition for a strict local minimum.

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Reflect and continue

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