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Convex KKT conditions certify a global minimum and an outward normal

Statement

Let $m \geq 0$ be an integer. Let $U \subseteq R^{n}$ be open and convex, and suppose the objective $f : U \to R$ and inequalities $g_{i} : U \to R$ $(1 \leq i \leq m)$ are differentiable and convex. Let the equalities be $A x = b$ and define the feasible set $C = {x \in U : A x = b, g_{i} (x) \leq 0 (1 \leq i \leq m)}$ . If a feasible tuple $(x_{*}, ν, λ)$ satisfies the KKT conditions with $λ \geq 0$ , then $x_{*}$ is a global minimizer over the feasible set $C$ . Moreover, $- \nabla f (x_{*}) \in N_{C} (x_{*})$ , where the normal cone has the convention of Scaled KKT residual and convex normal cone.

Facts & Assumptions

Given: The convex program, feasible KKT tuple, and normal-cone definition.

[F1]

For differentiable convex $φ$ on an open convex set, $φ (y) \geq φ (x) + \nabla φ (x)^{T} (y - x)$ (Differentiable convex functions are characterized by the gradient inequality).

[F2]

The normal cone is $N_{C} (x) = {v : ⟨ v, y - x ⟩ \leq 0 for every y \in C}$ (Scaled KKT residual and convex normal cone).

Proof

1.1givenF1algebra

Fix any feasible $y$ and put $d = y - x_{*}$ . Then $A d = 0$ . KKT stationarity gives $\nabla f (x_{*})^{T} d = - \sum_{i} λ_{i} \nabla g_{i} (x_{*})^{T} d$ . If $λ_{i} = 0$ its term is zero. If $λ_{i} > 0$ , complementarity gives $g_{i} (x_{*}) = 0$ , and [F1] applied to $g_{i}$ gives $\nabla g_{i} (x_{*})^{T} d \leq g_{i} (y) \leq 0$ . Hence $\nabla f (x_{*})^{T} d \geq 0$ .

2.1step 1.1F1F2algebra∎

Applying [F1] to $f$ yields $f (y) \geq f (x_{*}) + \nabla f (x_{*})^{T} d \geq f (x_{*})$ . Since $y$ was arbitrary, $x_{*}$ is globally minimizing. Step 1.1 also gives $⟨ - \nabla f (x_{*}), y - x_{*} ⟩ \leq 0$ for every feasible $y$ , which is exactly $- \nabla f (x_{*}) \in N_{C} (x_{*})$ by [F2].

Depends on

Used by

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Sources

Boyd and Vandenberghe, Convex Optimization (standard reference, not scraped)
Nocedal and Wright, Numerical Optimization, second edition (standard reference, not scraped)