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An unconstrained KKT point that is a saddle

Example

With no constraints, let $f (x_{1}, x_{2}) = x_{1}^{2} - x_{2}^{2}$ . The origin satisfies the KKT stationarity equation, but is a saddle point rather than a local minimum.

Facts & Assumptions

Given: The unconstrained $C^{2}$ objective $f (x_{1}, x_{2}) = x_{1}^{2} - x_{2}^{2}$ and $x_{*} = (0, 0)$ .

[F1]

With no constraints, KKT is $\nabla f (x_{*}) = 0$ and the critical cone is all of $R^{2}$ (Inequality constrained programs, tangent cones, constraint qualifications, and KKT).

[F2]

At a local minimizer satisfying LICQ, the Lagrangian Hessian is nonnegative on the critical cone; strict positivity on every nonzero critical direction is sufficient for a strict local minimum (Second-order necessity on the critical cone under LICQ, Strict second-order positivity on the critical cone gives a strict local minimum).

Verification

1.1givenF1algebra

The gradient is $\nabla f (x_{1}, x_{2}) = (2 x_{1}, - 2 x_{2})$ , so $\nabla f (0, 0) = 0$ and the origin is KKT. With no constraints, $L_{C} (0) = R^{2}$ and $K (0) = R^{2}$ . The Hessian is $(\begin{array}{cc} 2 & 0 \\ 0 & - 2 \end{array})$ , whose quadratic form at $d = (0, 1)$ is $- 2$ .

2.1givenF2step 1.1algebra

For every $t \neq 0$ , $f (t, 0) = t^{2} > 0 = f (0, 0)$ , while $f (0, t) = - t^{2} < 0 = f (0, 0)$ . Both points approach the origin as $t \to 0$ , so the origin is neither a local minimum nor a local maximum. The negative critical curvature also violates the necessary condition in [F2], and strict positivity required by its sufficient condition fails.

3.1step 1.1step 2.1∎

Thus stationarity alone is not an optimality certificate: this KKT point is a saddle.

Depends on

Used by

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Dependency tree · two levels

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Sources

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