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Second-order necessity on the critical cone under LICQ

Statement

For the $C^{2}$ program of Inequality constrained programs, tangent cones, constraint qualifications, and KKT, let $x_{*}$ be a local minimizer satisfying LICQ, and let $(ν, λ)$ be its KKT multipliers. Then

d^{T} \nabla_{x x}^{2} L (x_{*}, ν, λ) d \geq 0 for every d \in K (x_{*}, ν, λ) .

Facts & Assumptions

Given: The $C^{2}$ program, local minimizer $x_{*}$ , LICQ, and a critical direction $d$ .

[F1]

LICQ gives existence of KKT multipliers at a local minimizer (LICQ implies MFCQ and makes KKT multipliers unique).

[F2]

A critical direction lies in the linearized cone and satisfies $D g_{i} (x_{*}) d = 0$ for every active $i$ with $λ_{i} > 0$ (Critical directions are tangent to every positively weighted active inequality).

[F3]

A regular level set realizes any second-order jet satisfying $D H (x_{*}) d = 0$ and $D H (x_{*}) z + D^{2} H (x_{*}) [d, d] = 0$ (A regular level set realizes every compatible second-order jet).

[F4]

For a $C^{2}$ scalar function, $f (x_{*} + s) = f (x_{*}) + \nabla f (x_{*})^{T} s + \frac{1}{2} s^{T} \nabla^{2} f (x_{*}) s + o (∥ s ∥^{2})$ (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ ).

Proof

1.1givenF1F2construct

If $d = 0$ the result is immediate, so assume $d \neq 0$ . Let $P = {i \in I (x_{*}) : λ_{i} > 0}$ and $Z = {i \in I (x_{*}) : λ_{i} = 0, D g_{i} (x_{*}) d = 0}$ . Put $H = (h, (g_{i})_{i \in P})$ . By LICQ the rows of $D H (x_{*})$ , together with the rows $D g_{i} (x_{*})$ for $i \in Z$ , are independent. Thus the combined finite-dimensional linear map is onto. Choose $z$ solving $D h (x_{*}) z + D^{2} h (x_{*}) [d, d] = 0, D g_{i} (x_{*}) z + D^{2} g_{i} (x_{*}) [d, d] = 0 (i \in P),$ and $D g_{i} (x_{*}) z + D^{2} g_{i} (x_{*}) [d, d] = - 1 (i \in Z) .$ The first two groups say $D H (x_{*}) z + D^{2} H (x_{*}) [d, d] = 0$ .

2.1step 1.1F2F3F4algebra

By [F3] there is a $C^{2}$ curve $γ$ with $H (γ (t)) = H (x_{*})$ , $γ (0) = x_{*}$ , $γ^{'} (0) = d$ , and $γ^{''} (0) = z$ . For $i \in P$ the corresponding active inequality is exactly fixed at zero. An active $i \notin P$ has $λ_{i} = 0$ ; by [F2], either $D g_{i} (x_{*}) d < 0$ , giving $g_{i} (γ (t)) = t D g_{i} (x_{*}) d + O (t^{2}) < 0$ , or $i \in Z$ , giving $g_{i} (γ (t)) = - t^{2} / 2 + o (t^{2}) < 0$ . Inactive inequalities remain strict by continuity. Hence $γ (t)$ is feasible for all sufficiently small $t > 0$ .

2.2F1F2step 1.1F4algebra

Taylor expansion [F4], criticality $\nabla f (x_{*})^{T} d = 0$ , and $γ (t) = x_{*} + t d + \frac{1}{2} t^{2} z + o (t^{2})$ give $f (γ (t)) - f (x_{*}) = \frac{1}{2} t^{2} (\nabla f (x_{*})^{T} z + d^{T} \nabla^{2} f (x_{*}) d) + o (t^{2}) .$ Stationarity gives $\nabla f (x_{*})^{T} z = - ν^{T} D h (x_{*}) z - \sum_{i \in P} λ_{i} D g_{i} (x_{*}) z$ . The jet equations in step 1.1 therefore make the coefficient in parentheses equal to $d^{T} \nabla_{x x}^{2} L (x_{*}, ν, λ) d$ .

3.1step 2.1step 2.2givenassume-contradischarge-contradiction∎

Local minimality gives $f (γ (t)) - f (x_{*}) \geq 0$ for small $t > 0$ . If the displayed Lagrangian quadratic form were negative, step 2.2 would make this difference negative for all sufficiently small positive $t$ , a contradiction. Thus the quadratic form is nonnegative, as claimed.

Depends on

Used by

An unconstrained KKT point that is a saddle Example

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nocedal and Wright, Numerical Optimization, second edition (standard reference, not scraped)
J. M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)