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Exact-Hessian SQP converges quadratically on the identified active KKT system

Statement

Let x∗ be a KKT point of a C2 nonlinear program, with multiplier μ∗=(ν∗,λ∗,I) and active set I=I(x∗). Assume LICQ, strict complementarity λ∗,i>0 for every i∈I, and the second-order sufficient condition dT∇xx2L(x∗,μ∗)d>0for every nonzero critical direction d. Assume also that the active KKT map FA has a locally Lipschitz Jacobian: on some ball, ∥DFA(z)−DFA(z′)∥2≤L∥z−z′∥2. Once the correct active set is used, exact arithmetic full steps of Full-step exact-Hessian SQP after active-set identification, started sufficiently close to (x∗,μ∗), are well-defined and converge quadratically to that pair. This is the local full-step Newton result; it does not assert acceptance by an L1 merit line search.

Facts & Assumptions

Given: The smooth program, a KKT pair, LICQ, strict complementarity, the stated second-order sufficient condition, a locally identified active set, and the locally Lipschitz Jacobian hypothesis.

[F1]

Inequality constrained programs, tangent cones, constraint qualifications, and KKT defines the KKT system, LICQ, active set, and critical cone in the g≤0 convention.

[F2]

The SQP KKT matrix is nonsingular under positive reduced curvature: if active rows are independent and the Lagrangian Hessian is positive on their null space, the active KKT Jacobian is nonsingular.

[F3]

Full-step exact-Hessian SQP after active-set identification solves the exact Newton equation for the active KKT map and takes the full step.

[F4]

Forcing residuals determine local inexact-Newton rates gives local quadratic convergence for a C1 root map with Lipschitz Jacobian and invertible root Jacobian when the Newton residual is zero.

[A1]

The program is C2, LICQ holds, all active inequality multipliers are strictly positive, the Lagrangian Hessian is positive on every nonzero critical direction, and the active KKT map has a locally Lipschitz Jacobian near the root.

Proof

1.1givenF1algebra

Let A stack equality rows and active inequality rows at x∗. For d∈ker⁡A, d is linearly feasible and has zero derivative on every active inequality. Conversely, if d is in the critical cone, stationarity gives ∇f(x∗)Td=−∑i∈Iλ∗,iDgi(x∗)d; each Dgi(x∗)d≤0, so equality to zero and strict positivity of every λ∗,i force each such derivative to vanish. Thus the critical cone is exactly ker⁡A. The stated second-order condition gives positive curvature on ker⁡A, and LICQ gives independent rows.

2.1givenA1F2F3F4step 1.1algebra∎

Define F(x,μ)=(∇xL(x,μ),cA(x)). Its derivative at the root is the active KKT block matrix, nonsingular by [F2] and step 1.1. By [A1], this derivative is locally Lipschitz on a small ball. The exact step of [F3] satisfies DF(xk,μk)sk+F(xk,μk)=0. Apply [F4] to the combined primal-dual variable: for a sufficiently close starting pair, all full steps are defined, remain local, and converge quadratically.

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