Classify stationary points without overclaiming

Try

At the origin, each of x²+y², x²−y² and −x²−y² has zero gradient. Yet the first has a strict minimum, the second a saddle and the third a strict maximum. A stationary point is a candidate, not a classification. Inspect the function along coordinate directions: the saddle increases along one axis and decreases along the other. This directional evidence already disproves minimality without solving any global optimization problem.

Predict

Predict whether a positive-semidefinite Hessian at a stationary point is sufficient for a local minimum. Compare x⁴ and x³ at zero. Both have zero first and second derivatives, but only x⁴ has a local minimum. The missing information lies in higher-order terms. The second-order test is conclusive with positive definiteness or a negative-curvature direction, but can be inconclusive when the quadratic term vanishes in some directions.

Experiment

For each quadratic, calculate the Hessian and evaluate dᵀHd in two perpendicular directions. The factors of two come from differentiation, not from an arbitrary convention.

Function	Hessian	Classification at 0
x²+2y²	diag(2,4)	Strict global minimum
x²−2y²	diag(2,−4)	Saddle
−x²−2y²	diag(−2,−4)	Strict global maximum

Then inspect f(x,y)=x²+y⁴. Its Hessian at zero has a zero eigenvalue, yet the function is positive away from zero. Explain why this is a failed converse to a sufficient test, not a contradiction of that test.

Understand

At an interior local minimum of differentiable f, every one-variable restriction t↦f(x⋆+td) has a local minimum at zero for sufficiently small positive and negative t. Its derivative is ∇f(x⋆)·d, which must vanish for every d. Thus the gradient is zero. Interiority matters: a constrained boundary minimum may have a nonzero gradient because one sign of a direction is infeasible.

For C² f, the same line restriction implies dᵀ∇²f(x⋆)d≥0 at a local minimum. Conversely, a zero gradient and positive-definite Hessian imply a strict local minimum. Taylor expansion gives a quadratic term bounded below by a positive multiple of ||h||², while the remainder is o(||h||²), too small to reverse the sign nearby. If a direction has strictly negative quadratic form at a stationary point, small steps along it decrease the function, ruling out a local minimum. These are local statements. To claim global minimality from a zero gradient, add convexity and apply the supporting-hyperplane inequality.

Apply

Consider f(x)=x⁴−2x². The derivative 4x(x²−1) gives stationary points −1,0,1. The second derivative 12x²−4 is positive at ±1 and negative at zero. Thus ±1 are strict local minima and zero is a local maximum. Completing the square gives f=(x²−1)²−1, proving that both minima are global with value −1. The derivative test classified candidates; a separate identity established the global bound.

Now add a small linear tilt εx. A numerical method may approach one of two different local minima, and their objective values need not agree. The old symmetric completion-of-square certificate no longer establishes equality. Recheck the changed model rather than assuming a previously valid optimality claim survives perturbation unchanged.

Continue

Use three labels carefully: stationary, locally minimizing and globally minimizing. A small numerical gradient is evidence of approximate stationarity only, unless additional curvature, feasibility and global-bound arguments are supplied. Continue to Linear programming geometry, where linear objectives and polyhedral constraints make global certificates particularly concrete. Boundary optima there will demonstrate why unconstrained zero-gradient rules are not the correct conditions for every problem.